BRAVO! You got it right!

Since we are

notparticularly interested in thetime, we can usev

^{2}= v_{i}^{2 + 2 a (x - xi)}one of our "Big Three Kinematics Equations" which will come in very, very handy as we study motion.

Remember, of course, that v is the "final velocity" or v is the velocity when the position is x. We are looking for the velocity at the

topso that means v = 0 and x = x_{top. xi is the initial position; xi = 0.Gravity is the only thing affecting this motion; so this is a free fall problem. That means}a = - g = - 9.8 m/s

^{2}That

minus signis important!Okay, we have all the pieces; let's put them together.

v

^{2}= v_{i}^{2 + 2 a (x - xi)}0 = (25 m/s)

^{2}+ 2 ( - 9.8 m/s^{2})( x - 0 )^{}Do a quick check of units!,

0 = (25 m/s)

^{2}+ 2 ( - 9.8 m/s^{2}) x2 ( 9.8 m/s

^{2 }) x = (25 m/s)^{2}x = ( 625/19.6) m

x = 31.9 m

(c) 2000, Doug Davis; all rights reserved.