Question

a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. to what height did the ball rise.

t=7.72 s

the answer is 73 m, but how do you get that answer

Answer 1

initial KEnergy=final PEnergy
1/2 m v^2=mgh

h= 1/2 *v^2/g=1/2 * 37.8^2/9.8 in my head about = 1600/20=80 meters, work it out accurately. 73 meters looks good

Answer 2

You can use the formula for Uniformly Accelerated Motion (UAM):
h = (v,o)t - (1/2)gt^2

Or you can use another formula, which does not use time:
v,f^2 - v,o^2 = 2gd
where
v,f = final velocity
v,o = initial velocity
g = acceleration due to gravity = 9.8 m/s^2
d = distance (in this case, it's the height)

Note that at the highest point, the ball does not move (v,f = 0). We're solving for d. Substituting to the second equation:
0 - (37.8)^2 = 2(-9.8)(d)
-1428.84 = -19.6*d
d = 1428.84 / 19.6
d = 72.9 m

The negative sign of g indicates the direction (since acceleration is a vector quantity).
Hope this helps :)