**Ch4: 2, 5, 13, 17, 29, 37, 39, 47, 55, 65, 66**

**Questions 1, 4, 8, 13, 17, 30**

Additional problems from Serway's fourth edition

(4 ed) 4.2 A ball is thrown horizontally from the top
of a building 35 m high. The ball strikes the ground at a point 80 m from
the base of the building.

**Find the following:**

**(a) the time the ball is in flight,**

**(b) its initial velocity, and**

**(c) the x and y components of velocity just before the
ball strikes the ground.**

(4 ed) 4.3 Superman is flying at treetop level near Paris
when he sees the Eiffel Tower elevator start to fall (the cable snapped).
His x-ray vision tells him Lois Lane is inside. If Superman is 1.00 km
away from the tower and the elevator falls from a height of 240 m, how
long does have to save Lois, and what must his average speed be?** **

**(4 ed) 4.4 A tire 0.500 m in radius rotates at a constant
rate of 200 rev/min (rpm). Find the speed and acceleration of a small
stone lodged in the tread on the outer edge of the tire.**

(4 ed) 4.5 At t = 0 a particle leaves the origin with
a velocity of 6.00 m/s in the positive y direction. Its acceleration is
given by

**When the particle reaches its maximum y coordinate, its
y component of velocity is zero. **

**At this instant, find **

**(a) the velocity of the particle and**

**(b) its x and y coordinates.**

(4 ed) 4.7 After delivering his toys in the usual manner,
Santa decides to have some fun and slide down an icy roof, as in Figure
P4.78. He starts from rest at the top of the roof, which is 8.00 m in
length, and accelerates at the rate of 5.00 m/s^{2}. The edge
of the roof is 6.00 m above a soft snow bank, which Santa lands on.

**Find the following:**

**(a) Santa's velocity components when he reaches the snow
bank,**

**(b) the total time he is in motion, and****(c) the distance
d between the house and the point where he lands in the snow.**

Conceptual Questions

Q4.1 Can an object accelerte if its speed is constant? Can an object accelerate if its velocity is constant?Acceleration is the change in

velocitydivided by the change in time. So if thevelocityisconstantthe change invelocityis zero and the acceleration iszero. However, thedirectionof the velocity can change even if its magnitude -- thespeed-- is constant. Uniform Circular Motion (UCM) is a good example of nonzero acceleration withconstant speed.

Q4.4 Describe a situation in which the velocity of a particle is always perpendicular to the position vector.Uniform Circular Motion (UCM) is a good example of this:

Q4.8 A rock is dropped at the same instant that a ball at the same initial elevation is thrown horizontally. Which will have the greater speed when it reaches ground level?They will

bothhave thesame verticalvelocity component. However the one thrown horizontally willalsohave a horizontal velocity component. Therefore, the rock thrown horizontally will have thegreaterspeed.

Q4.13 At the end of its arc, the velocity of a pendulum is zero. Is it acceleration also zero at this point?

No!If its acceleration were zero then its velocity wouldremainzero and it would just stop, suspended in mid-air. Just before it stopped, the pendulum's velocity was in one direction and a moment later its velocity will be in the opposite direction.

Q4.17 A baseball is thrown with an initial velocity of (10i+ 15j) m/s. When it reaches the top of its trajectory, what are

(a) its velocity andAt the top, its

verticalvelocity component is zero. Throughout its flight, thehorizontalvelocity component remains constant. Therefore, its total velocity at the top is

v_{top}= ( 10i+ 0j) m/s

v_{top}= ( 10i) m/s

(b) its acceleration?Throughout the flight, the acceleration remains constant,

a=g= - (9.8 m/s^{2})j

Neglect air resistance

Q4.20 A projectile is fired at an angle of 30other^{o}from the horizontal with some initial speed. Firing at whatprojectile angle results in the same range if the initial speed is the same in both cases? Neglect air resistance.Any angle and 90

^{o}minus that angle have the same range. Therefore, the range is the same for 30^{o}and for 60^{o}.

Problems from the current (5th) edition of Serway and Beichner.

4.2 Suppose that the position vector for a particle is given asr(t) = x(t)i+ y(t)j

withx(t) = a t + bandy(t) = ct^{2}+ d,where a = 1.00 m/s, b = 1.00 m, c = 0.125 m/s^{2}, and d = 1.00 m.

(a) Calculate the average velocity during the time interval from t = 2.00 s to t = 4.00 s.v_{avg}=r/ t

r=r(4 s) -r(2s)

r(4 s) = x(4 s)i+ y(4 s)j

r(4 s) = [ a (4 ) + b ]i+ [ c (4)^{2}+ d ]j

r(4 s) = [ (1) (4) + 1 ]i+ [ (0.125) (4)^{2}+ 1 ]j

r(4 s) = [ 5 ]i+ [ 3 ]j

r(2 s) = x(2 s)i+ y(2 s)j

r(2 s) = [ a (2 ) + b ]i+ [ c (2)^{2}+ d ]j

r(2 s) = [ (1) (2) + 1 ]i+ [ (0.125) (2)^{2}+ 1 ]j

r(4 s) = [ 3 ]i+ [ 1.5 ]j

r=r(4 s) -r(2s)=[ 5i+ 3j] - [ 3i+ 1.5j]

r= 2i+ 1.5j

v_{avg}=r/ t

v_{avg}= [2i+ 1.5j] m / 2 s

v_{avg}= ( 1i+ 0.75j) m/s

(b) Determine the velocity and the speed at t = 2.00 s.v= dr/ dt = [ dx / dt ]i+ [ dy / dt ]j

v= [ d( a t + b) / dt ]i+ [ d( ct^{2}+ d ) / dt ]j

v= ai+ 2 c tj

v= 1i+ (2) (0.125) (2)j

v=i+ 0.5j

v= (i+ 0.5j) m/s for t = 2 sv = SQRT[ v

_{x}^{2}+ v_{y}^{2}] = SQRT[ (1)^{2}+ (0.5)^{2}]^{m}/_{s}= SQRT[ 1.25 ]^{m}/_{s}= 1.12^{m}/_{s}

4.5 At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of

v(3_{o}=i -2j)m/s at the origin(i.e., r(9_{o}= 0). At t = 3 s, its velocity is given by v =i +7j)m/s. Find

(a) the acceleration of the particle anda=v/ t

v=v-_{f}v_{i}

v=v(3 s) -v(0)

v= [ 9i+ 7j] - [3i- 2j]

v= 6i+ 9j

a=v/ t= [ 6i+ 9j] / 3

a= [ 2i+ 3j] m / s^{2}

(b) its coordinates at any time t.r=r_{o}+v_{o}t +^{1}/_{2}at^{2}

r= [ 0i+ 0j] + [ 3i -2j] t +^{1}/_{2}[ 2i+ 3j] t^{2}

r= [ [ 3 t + t^{2}]i+ [ - 2 t + (^{3}/_{2}) t^{2}]j] mx = ( 3 t + t

^{2}) my = [ - 2 t + (

^{3}/_{2}) t^{2}] m

4.13 An artillery shell is fired with an initial velocity of 300 m/s at 55.0^{o}above the horizontal. It explodes on a mountainside 42.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?v

_{x}= v cos = (300 m/s) cos 55^{o}v

_{x}= (300 m/s) (0,574) = 172,1 m/sv

_{y}= v sin = (300 m/s) sin 55^{o}v

_{y}= (300 m/s) (0,819) = 245.7 m/sFor motion with constant acceleration, we know

s = s

_{o}+ v_{o}t + (1/2) a t^{2}Along the horizontal, x-axis, we might write this as

x = x

_{o}+ v_{xo}t + (1/2) a_{x}t^{2}Measuring distances relative to the firing point means

x

_{o}= 0We know that

a

_{x}= 0or

v

_{x}= v_{xo}= constantv

_{x}= 172.1 m/sx = (172.1 m/s) (42 s)

x = 72 282 m = 72.3 kmAlong the vertical, y-axis, we might write this as

y = y

_{o}+ v_{yo}t + (1/2) a_{y}t^{2}Measuring distances relative to the firing point means

y

_{o}= 0We know that

a

_{y}= - g = - 9.8 m/s^{2}y = 0 + (245.7 m/s) t - (1/2) (9.8 m/s

^{2}) t^{2}y = 0 + (245.7 m/s) (42 s) - (1/2) (9.8 m/s

^{2}) (42 s)^{2}

y = 1 675.8 m

4.17 A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired?

v

_{x}= v cosv

_{x}= (1 000 m/s) cosv

_{y}= v sinv

_{oy}= (1 000 m/s) sina

_{x}= 0v

_{x}= constantx = v

_{x}ta

_{y}= - g = - 9.8 m/s^{2}y = y

_{o}+ v_{yo}t + (1/2) a_{y}t^{2}y =v

_{yo}t - (1/2) g_{}t^{2}y = [(1 000 m/s) sin] t - (1/2)(9.8 m/s

^{2}) t^{2}y = [(1 000) sin] t - (1/2)(9.8) t

^{2}t = x / v

_{x}t = 2 000 m /[(1 000 m/s) cos ]

_{}t = 2 000/[(1 000) cos ]

t = 2/cos

y = [(1 000) sin] t - (1/2)(9.8) t

^{2}y = [(1 000) sin ] (2/cos ) - (1/2)(9.8) (2/cos )

^{2}y = [(2 000) (sin /cos )] - (19.6/cos

^{2})^{}800 = [(2 000) (sin /cos )] - (19.6/cos

^{2})800 cos

^{2}= 2 000 sin cos - 19.619.6 + 800 cos

^{2}= 2 000 sin cos19.6 + 800 cos

^{2}= 2 000 [SQRT(1-cos^{2})] cosNow square both sides for

384 + 31 360 cos

^{2}+ 640 000 cos^{4}= 4 000 000 cos^{2}+ 4 000 000 cos^{4}4 640 000 cos

^{4}- 3 968 640 cos^{2}+ 384 = 0cos

^{2}= 0.855 or 9.676 x 10^{ - 5}cos = 0.925 or 9.837 x 10

^{ - 3}= 22.4

^{o}or 89.4^{o}This was a rather

messyquadratic equation! But the whole problem was fairly straightforward. Just a bit messy with the unusually large numbers. But that's okay!

4.29 Young David, who slew Goliath, experimented with slings before tackling the giant. He found that with a sling of length 0.60 m, he could revolve the sling at the rate of 8.0 rev/s(Wow! that'sreallyfast!). If he increased the length to 0.90 m, he could revolve the sling only 6.0 times per second.

(a) Which rate of rotation gives the larger linear speed?v = rv

_{6}= (0.6 m) (8^{rev}/_{s}) ( 2/_{rev}) = 30.16 m/sv

_{9}= (0.9 m) (6^{rev}/_{s}) ( 2/_{rev}) = 33.93 m/sRotating it with a radius of

0.90 mand an angular velocity of6 rev/sgives thegreater linear speed.

(b) What is the centripetal acceleration at 8.0 rev/s?a_{c }= v^{2}/ra

_{c }= (30.16 m/s)^{2}/ 0.6 ma

_{c }= 1 516 m/s^{2}

(c) What is the centripetal acceleration at 6.0 rev/s?a_{c }= v^{2}/ra

_{c }= (33.93 m/s)^{2}/ 0.9 ma

_{c }= 1 279 m/s^{2}

4.37 A river has a steady speed of 0.50 m/s. A student swims upstream ad istance of 1.0 km and swims back to the starting point. If the student can swim at a speed of 1.2 m/s in still water, how long does the trip take? Compare this with the time the trip would take if the water were still.v

_{upstream}= (1.2 - 0.5) m/s = 0.7 m/st

_{upstream}= 1 000 m / (0.7 m/s) = 1 429 sv

_{downstream}= (1.2 + 0.5) m/s = 1.7 m/st

_{downstream}= 1 000 m / (1.7 m/s) = 588 st

_{total}= 1 429 s + 588 s = 2 017 sNow, if the trip were done in

still waterwe would simply havet

_{still}= 2 000 m / (1.2 m/s) = 1 667 s

4.39 The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.

V_{pe}= velocity ofplane wrtearth

V_{pa}= velocity ofplane wrtair

Vae = velocity ofair wrtearth

V_{pe}is the hypotenuse of a right triangle, so we can readily find its magnitude by

Now we need the

directionof the velocityV_{pe}, the velocity of the plane with respect to (wrt) the earth. We have often measured directions relative to the x-axis. So we could specify the angle shown in the diagram here.However, directions in aviation are usually given relative to North. So we will find the angle shown here.

tan =

^{opp}/_{adj}=^{150}/_{30}= 5= 78.7

^{o}Or, we might be quite specific and state

v_{pe}= 153^{ km}/_{h}at 78.7^{o}West of North

4.47 A projectile is fired up an incline (with incline angle ) with an initial seed v_{i}, at an angle_{i}, with respect to the horizontal (_{i}> ) as shown in Figure P4.47.(a) Show that the projectile travels a distance d up the incline, where

d = [ 2 v

_{i}^{2}cos_{i }sin (_{i }- ) ] / [g cos^{2}](b) For what value of

_{i }is d a maximum, and what is that maximum value of d?Some may consider

trigan art form in its own right. But Physicists -- and other scientists and engineers -- view trig -- and math in gen'l -- as a very usefultool. ThePhysicsof this problem is not difficult. But it soon becomes a rather involvedtrigproblem.Now that we have an expression for d, we can find the

maximumvalue of d by taking the derivative wrt angle and setting that equal to zero and solving for the angle and then evaluating d for that value of . [[ Yes, that's long and involved! But that's just life in Science or life in Engineering or life in Technology. Ya gotta do it! ]]Can we

checkthis answer? Well, maybe or just a little. When the incline angle is zero, we're back to a horizontal range problem. This expression reduces toR

_{max}_{}= v^{2}/gand that

isthe value or expression we expect.

4.55 A boy can throw a ball a maximum horizontal distance of 40.0 m on a level field. How far can he throw the same ball vertically upward? Assume that his muscles give the ball the same speed in each case.We have derived the Range Equation,X

_{max}= v_{o}^{2}sin 2 / gThis range has its maximum value (X

_{max,max}?) for = 45^{o}. From this maximum value of the range, we can determine the initial speed v_{o},X

_{max,max}= v_{o}^{2}sin (2 x 45^{o}) / g = v_{o}^{2}/ g = v_{o}^{2}/ (9.8 m/s^{2}) = 40 mv

_{o}^{2}= (9.8 m/s^{2}) (40 m) = 392 m^{2}/ s^{2}v

_{o}= 19.8 m / sWith this initial speed, what is the maximum height possible?

In the text, the maximum height equation is developed, Equation 4.17, on page 79,

h = v

_{o}^{2}sin^{2}/ 2 gOf course, this has its maximum value for = 90

^{o}so sin=1h

_{max}= v_{o}^{2}/ 2 gh

_{max}= (19.8 m / s)^{2}/ 2 (9.8 m/s^{2})h

_{max}= 20 m

4.65 A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37^{o}below the horizontal. The negligent driver leaves the car in neutral and the parking brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.0 m/s^{2}, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean.Find

(a) the speed of the car when it reaches the edge of the cliff and the time it takes to get there.

(b) the velocity of the car when it lands in the ocean,

(c) the total time the car is in motion, and

(d) the position of the car when it lands in the ocean, relative to the base of the cliff.

(a) the speed of the car when it reaches the edge of the cliff and the time it takes to get there.

This is straight-line motion with constant acceleration:

v

^{2}= v_{o}^{2}+ 2 a sv

^{2}= 0 + 2 (4.0 m/s^{2}) (50 m)v

^{2}= 400 m^{2}/s^{2}v = 20 m/s

With an acceleration of 4.0 m/s

^{2}, how long does it take to go from rest to 20 m/s?v = v

_{o}+ a t20 m/s = 0 + (4.0 m/s

^{2}) tt = 5 s

(b) the velocity of the car when it lands in the ocean,

How long is the car in the air after leaving the cliff, before it hits the ocean?

y = y

_{o}+ v_{yo}t + (1/2) a_{y}t^{2}y = y

_{o}+ v_{yo}t - (1/2) g_{}t^{2}0 = 30 m + ( - 12 m/s) t - (1/2) (9.8 m/s

^{2}) t^{2}Check the units. That is, do a

dimensional analysisof this equation. Every term is adistance, measured in m. That's good. That give us confidence that the equation is correct (of course, that does not guarantee that it is correct). If we measure time t in units of seconds, then every term has only the unit of m in it and we can divide by that to have0 = 30 - 12 t - 4.9 t

^{2}4.9 t

^{2}+ 12 t - 30 = 0t = 1.53 s

If we used the very reasonable approximation that g = 10 m/s

^{2}, this would be t = 1.5 s. In either case, there is also a solution for t < 0. While that satisfies the mathematics, it clearly does not satisfy the physics. Negative times are simply extraneous solutions.Now, after falling for t = 1.53 s, what is the car's vertical velocity component?

v

_{y}= v_{yo}+ a_{y}tv

_{y}= - 12 m/s + ( - 9.8 m/s^{2}) (1.53 s)v

_{y}= - 27 m/sRemember, of course, that the negative sign means it is falling

down.The horizontal component of the velocity remains constant,

v

_{x}= 16 m/sv = SQRT (v

_{x}^{2}+ v_{y}^{2})v = 31.4 m/s

That is the

speedof the car. To find thevelocity, we also need thedirection,tan =

^{opp}/_{adj}= 27/16 = 1.69= 59.3

^{o}Notice that the "opposite side" was used as just 27, not - 27. This gives us the angle as shown in the diagram. We can then describe that as 59.3

^{o}below the horizontal or as = - 59.3^{o}. That choice is up to you. Just be sure you know -- and communicate -- what the angle is.

(c) the total time the car is in motion,

Rolling down the incline took 5 s,

t

_{1}= 5.0 sIt is in the air, falling from the edge of the cliff into the ocean, for 1.53 s,

t

_{2}= 1.53 sSo the total time is just the sum of these two times,

t

_{tot}= 6.53 s

(d) the position of the car when it lands in the ocean, relative to the base of the cliff.

During this 1.53 s of free fall, the horizontal velocity remains constant a 16 m/s,

x = v

_{x}tx = (16 m/s)(1.53 s)

x = 24.5 m

4.66 The determined Wyle E Coyote is out once more to try to capture the elusive Road Runner. The Coyote wears a pair of Acme jet-powered rollerblades, which provide a constant horizontal acceleration of 15.0 m/s^{2}. See Figure P4.66. The Coyote starts off at rest 70.0 m from the edge of a cliff at the instant the RoadRunner zips past him in the direction of the cliff.(a) If the RoadRunner moves with constant speed, determine the minimum speed he must have to reach the cliff before the Coyote.

At the brink of the cliff, the RoadRunner escapes by making a sudden turn, while the Coyote continues straight ahead.

(b) If the cliff is 100 m above the floor of a canyon, detemine where the Coyote lands in the canyon (assume his skates remain horizontal and continue to operate when he is in "flight").

(c) Determine the components of the Coyote's impact velocities.

[[ Yes, this, too, is long and involved; but I hope it is

funtoo! ]]

Please do this on your own and look at this solution only after you have done it yourself!

(a) If the RoadRunner moves with constant speed, determine the minimum speed he must have to reach the cliff before the Coyote.

Since we know the Coyote's acceleration, we can find his speed at the edge of the cliff and then the amount of time req'd to reach that speed,

v

^{2}= v_{o}^{2}+ 2 a sv

^{2}= 0 + 2 (15.0 m/s^{2}) (70 m)v

^{2}= 2100 m^{2}/s^{2}v = 45.8 m/s

With an acceleration of 15.0 m/s

^{2}, how long does it take the Coyote go from rest to 45.8 m/s?v = v

_{o}+ a t45.8 m/s = 0 + (15.0 m/s

^{2}) tt = 3.05 s

This means the Road Runner has 3.05 s to travel this distance of 70.0 m,

v

_{RR}= 70 m/3.05 sv

_{RR}= 22.9 m/s

(b) If the cliff is 100 m above the floor of a canyon, detemine where the Coyote lands in the canyon (assume his skates remain horizontal and continue to operate when he is in "flight").

How long is the Coyote in the air before smacking into the canyon floor? His initial vertical velocity is zero, v

_{yo}= 0, so his vertical motion is ordinary free fall,y = y

_{o}+ v_{yo}t + (1/2) a_{y}t^{2}y = y

_{o}+ v_{yo}t - (1/2) g_{}t^{2}0 = 100 m + 0 - (1/2) (9.8 m/s

^{2}) t^{2}A quick check of units -- or dimensional analysis -- shows us that each term in this equation is a distance and if we measure time t in seconds we can write this equations as

0 = 100 + 0 - 4.9 t

^{2}4.9 t

^{2}= 100t

^{2}= 20.41t = 4.52 s

How far does the Coyote move horizontally during this 4.52 s? He leaves the cliff with an initial horizontal velocity of v

_{xo}= 45.8 m/s and continues to have a horizontal acceleration of a_{x}= 12 m/s^{2}during this 4.52 s,x = x

_{o}+ v_{xo}t + (1/2) a_{x}t^{2}x = 0 + (45.8 m/s)(4.53 s) + (1/2)(12 m/s

^{2})(4.53 s)^{2}x = 330.6 m

(c) Determine the components of the Coyote's impact velocities.

v

_{x}= v_{xo}+ a_{x}tv

_{x}= 45.8 m/s + (12 m/s^{2})(4.53 s)v

_{x}= 100 m/sv

_{y}= v_{yo}+ a_{y}tv

_{y}= v_{yo}- g_{}tv

_{y}= 0 - (9.8 m/s^{2})(4.53 s)v

_{y}= - 44.4 m/s

Solutions to additional problems from Serway's fourth edition.

(4 ed) 4.1 Jimmy is at the base of a hill, while Billy is 30 m up the hill. Jimmy is at the origin of an xy coordinate system, and the line that follows the slope of the hill is given by the equation, y = 0.4 x, as shown in Figure P4.10. If Jimmy throws an apple to Billy at an angle of 50^{o}with respect to the horizontal, with what speed must he throw the apple if it is to reach Billy? v_{xo}= v_{o}cos 50^{o}= 0.643 v_{o}v

_{yo}= v_{o}sin 50^{o}= 0.766 v_{o}For the hill, = tan

^{-1}0.4 = 21.8^{o}r

_{Billy}= 30 mx

_{Billy}= (30 m) (cos 21.8^{o}) = (30 m)(0.928) = 27.85 my

_{Billy}= (30 m) (sin 21.8^{o}) = (30 m)(0.371) = 11.14 mx

_{apple}= v_{xo}ty

_{apple}= v_{yo}t + (^{1}/_{2}) a_{y}t^{2}= v_{yo}t - (^{1}/_{2}) g t^{2}For some time t, we require x

_{apple}= x_{Billy}and y_{apple}= y_{Billy}x

_{apple}= x_{Billy}v

_{xo}t = 27.85 m0.643 v

_{o}t = 27.85 mt = (27.85 m) / (0.643 v

_{o})t = 43.31 / v

_{o}Now we

usethis value of the time iny

_{apple}= y_{Billy}v

_{yo}t - (^{1}/_{2}) g t^{2}= 11.14 m[ 0.766 v

_{o}] [ 43.31 / v_{o}] - (^{1}/_{2}) [ 9.8 ] [ 43.31 / v_{o}]^{2}= 11.14[ 0.766 v

_{o}] [ 43.31 / v_{o}] - (^{1}/_{2}) [ 9.8 ] [ 43.31 / v_{o}]^{2}= 11.1433.18 - 9191 / v

_{o}^{2}= 11.149191 / v

_{o}^{2}= 22.04v

_{o}^{2}= 9191 / 22.04 = 417v

_{o}= 20.4 m/s

v_{o}= 20.4 m/sNow, let's

checkthis to ensure it iscorrect:t = 43.31 / v

_{o}= (43.31 / 20.4) s = 2.12 sx

_{apple}= v_{xo}t = 0.643 v_{o}t = (0.643) (20.4 m/s) (2.12 s) = 27.8 m = x_{Billy}y

_{apple}= v_{yo}t - (^{1}/_{2}) g t^{2}= 0.766 v_{o}t - (^{1}/_{2}) (9.8) t^{2}= ? = y_{Billy}y

_{apple}= (0.766) (20.4) (2.12) - (^{1}/_{2}) (9.8) (2.12)^{2}== 33.16 - 22.02 = 11.1 m = y_{Billy}

(4 ed) 4.2 A ball is thrown horizontally from the top of a building 35 m high. The ball strikes the ground at a point 80 m from the base of the building.^{}

Find the following:

(a) the time the ball is in flight,y = y_{o}+ v_{yo}t + (^{1}/_{2}) a_{y}t^{2}We are free to choose the origin of or coordinate system wherever we want. This time, let's choose the base of the building, so y

_{o}= 35 m and y_{f }= 0.y = 35 + 0 + (

^{1}/_{2}) ( - g) t^{2}y = 35 + 0 - 4.9 t

^{2}= 0 = y_{f}4.9 t

^{2}= 35t

^{2}= 7.14t = 2.67 s

(b) its initial velocity, andx = xo + vxo t80 m = 0 + v

_{xo}(2.67 s)v

_{xo}= 80 m / 2.67 sv

_{xo}= 29.9 m/s

(c) the x and y components of velocity just before the ball strikes the ground.v_{x}= v_{xo}= 29.9 m/s = constantv

_{y}= v_{yo}+ a_{y}tv

_{y}= 0 + ( - 9.8 m/s^{2}) ( 2.67 s )v

_{y}= - 26.2 m/s

(4 ed) 4.3 Superman is flying at treetop level near Paris when he sees the Eiffel Tower elevator start to fall (the cable snapped). His x-ray vision tells him Lois Lane is inside. If Superman is 1.00 km away from the tower and the elevator falls from a height of 240 m, how long does have to save Lois, and what must his average speed be?y = y_{o}+ v_{yo}t + (^{1}/_{2}) a_{y}t^{2}We are also free to choose the direction of the positive y-axis. This time, since everything happens going down, let's choose

downas positive. Then a_{y}= g = 9.8 m/s^{2}and the ground is located at y = 240 m from the initial position of the elevator, y_{o}= 0.y = 0 + 0 t + (

^{1}/_{2}) ( 9.8 ) t^{2}= 240 m4.9 t

^{2}= 240t

^{2}= 49.98t = 7.0 s

That is

how longthe elevator falls before hitting the ground. During that time, Superman must travel 1.00 km = 1 000 m; therefore, his average speed must bev = x / t = 1000 m / 7 s

v = 143 m/s

We could change this speed to km / hr for a better understanding of how fast this is,

v = 143

^{m}/_{s}[^{km}/_{1 000 m}] [^{3 600 s }/_{h}] = 514^{km}/_{h}514 km/h is 314 mi/h

(4 ed) 4.4 A tire 0.500 m in radius rotates at a constant rate of 200 rev/min (rpm). Find the speed and acceleration of a small stone lodged in the tread on the outer edge of the tire.v = rv = r = (0.500 m) ( 200

^{rev}/_{min}) ( 2/_{rev}) (^{1 min }/_{60 s}) = 10.5 m / sa

_{c }= v^{2}/ ra

_{c }= ( 10.5 m/s )^{2}/ 0.500 ma

_{c }= 220.5 m / s^{2}

(4 ed) 4.5 At t = 0 a particle leaves the origin with a velocity of 6.00 m/s in the positive y direction. Its acceleration is given bya= ( 2.00i- 3.00j) m/s^{2}

When the particle reaches its maximum y coordinate, its y component of velocity is zero.

At this instant, find

(a) the velocity of the particle andv_{o}= 6.00jm/sv

_{xo}= 0v

_{yo}= 6.00 m/s

r_{o}= 0v

_{y}= v_{yo}+ a_{y}tv

_{y}= 6 + ( - 3 ) t = 0t = 2 s

v

_{x}= v_{xo}+ a_{x}tv

_{x}= 0 + (2 m/s^{2 }) (2 s) = 4 m/s

(b) its x and y coordinates.x = x_{o}+ v_{xo}t + (^{1}/_{2}) a_{x}t^{2}x = 0 + 0 + (

^{1}/_{2}) (2 m/s^{2}) (2 s)^{2}x = 4 m

y = y

_{o}+ v_{yo}t + (^{1}/_{2}) a_{y}t^{2}y = 0 + (6 m/s) (2 s) + (

^{1}/_{2}) ( - 3 m/s^{2}) (2 s)^{2}y = 6 m

(4 ed) 4.6A football is thrown toward a receiver with an initial speed of 20.0 m/s at an angle of 30.0^{o}above the horizontal. At that instant, the receiver is 20.0 m from the quaterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?Where is the football going? That is, how far does it move horizontally by the time it gets back to the vertical level at which it was thrown? We need the initial components of the velocity (or the components of the initial velocity).v

_{xo}= v_{o}cos = ( 20 m/s ) cos 30^{o}= (20 m/s ) ( 0.866 ) = 17.32 m/sv

_{yo}= v_{o}sin = ( 20 m/s ) sin 30^{o}= (20 m/s ) ( 0.500 ) = 10.00 m/sSet y = y

_{o}+ v_{yo}t - (^{1}/_{2}) g t^{2}= y_{o}v

_{yo}t - (^{1}/_{2}) g t^{2}= 0[ v

_{yo}- (^{1}/_{2}) g t ] t = 0Either t = 0, which is "true" but "uninteresting", or

t = 2 v

_{yo}/ gt = 2 ( 10.00 m/s ) / (9.8 m/s

^{2})t = 2.04 s

Now we know

how longthe football is in the air.How fardoes it travel in this time?x = v

_{xo}tx = (17.32 m/s ) (2.04 s)

x = 35.3 m

At the time the quarterback throws the football, the receiver is 20 m away. During the next 2.04 s, he must run

awayfrom the quarterback to get to a distance of 35.3 m when the football arrives. The receiver isalready20 m away so he must run an additional 15.3 m in that time of 2.04 s. Therefore he needs to run with a constant (or average) speed ofv = 15.3 m / 2.04 s

v = 7.50 m/s

(4 ed) 4.7 After delivering his toys in the usual manner, Santa decides to have some fun and slide down an icy roof, as in Figure P4.78. He starts from rest at the top of the roof, which is 8.00 m in length, and accelerates at the rate of 5.00 m/s^{2}. The edge of the roof is 6.00 m above a soft snow bank, which Santa lands on.

Find the following:

(a) Santa's velocity components when he reaches the snow bank,While sliding along the icy roof for 8.00 m, Santa's acceleration, we are told, is 5.00 m/s^{2}For this motion along a straight line, we can use

v

^{2}= v_{o}^{2}+ 2 a ( s - s_{o})v

^{2}= 0 + 2 ( 5 m/s^{2}) ( 8 m )v

^{2}= 80 m^{2}/s^{2}v = 8.94 m/s

This is Santa's speed at the eave of the roof; call this v1.

v

_{1}= 8.94 m/sThe roof is inclined 37

^{o}, sov

_{x1}= v_{1}cos 37^{o}= (8.94 m/s) (0.80) = 7.16 m/sv

_{y1}= - v_{1}sin 37^{o}= - (8.94 m/s) (0.60) = - 5.37 m/sSanta reaches the snow bank 6.00 m below this point. Call this point x

_{1}= 0, y_{1}= 6.0 m and then Santa reaches the snow bank at y = 0. How long does this take? For what value of the time t does y = 0?y = y

_{1}+ v_{y1}t + (^{1}/_{2}) (- g) t^{2}= 06.0 m - (5.37 m/s) t - (4.9 m/s

^{2}) t^{2}= 06.0 - 5.37 t - 4.9 t

^{2}= 04.9 t

^{2}+ 5.37 t - 6.0 = 0Now we use the quadratic equation and find

t

_{1}= 0.69 st

_{2}= - 1.78 sOf course, we are interested only in the positive value,

t = 0.69 s

v

_{y}= v_{y1}+ a_{y}tv

_{y}= - 5.37 m/s - 9.8 m/s^{2}(0.69 s)v

_{y}= - 12.13 m/sv

_{x}= v_{x1}= 7.16 m/s = constant

(b) the total time he is in motion, andTo slide down the icy roof requires time t_{roof}. In finding Santa's speed as he comes off the roof, we did not find this time. We can find it now fromv = v

_{o}+ a t8.94 m/s = 0 + (5.0 m/s

^{2}) tt = t

_{roof }= 1.79 sAnd we just found that Santa is airborne for an additional time of t

_{air}t

_{air}= 0.69 st

_{tot}= t_{roof}+ t_{air}= 2.48 s

(c) the distance d between the house and the point where he lands in the snow.Once he leaves the roof, Santa'shorizontal speedremains constant,x = v

_{xo}tx = (7.16 m/s) (0.69 s)

x = 4.9 m

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