Posted by **Aly** on Sunday, December 1, 2013 at 10:08pm.

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

- Physics -
**bobpursley**, Sunday, December 1, 2013 at 10:20pm
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

- Physics -
**Jai**, Sunday, December 1, 2013 at 10:20pm
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 :)

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