# Physics

posted by .

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 -

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 -

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 :)

## Similar Questions

1. ### Physics

A batted baseball leaves the bat at an angle of 30.0 degrees about the horizontal and is caught by an outfilder 375 ft from home plate at the same height from which it left the bat. (a) what was the initial speed on the ball?
2. ### physics-momentum

A 0.145-kg baseball pitched horizontally at 39 m/s strikes a bat and is popped straight up to a height of 31 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] …
3. ### Physics

I solved part A but not part B: A batted baseball leaves the bat at an angle of 35.0 degrees above the horizontal and is caught by an outfielder 370 ft. from home plate at the same height from which it left the bat. Part A: What was …
4. ### physics

A 0.01-kg baseball traveling in a horizontal direction with a speed of 12 m/s hits a bat and is popped straight up with a speed of 15 m/s. (a) What is the change in momentum (magnitude and direction) of the baseball?
5. ### physics for scientists and engeneers I

A 0.145-kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to a height of 43 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] …
6. ### physics

A batted baseball leaves the bat at an angle of 26.0^\circ above the horizontal and is caught by an outfielder 355{\rm ft} from home plate at the same height from which it left the bat. What was the initial speed of the ball?
7. ### physics

A 0.145-kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to a height of 38 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] …
8. ### physics. again.

a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. at what time after being struck is the ball moving at 10 m/s upwards?
9. ### physics

a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. at what time is the ball 20 m above the ground?
10. ### physics

A batted baseball leaves the bat at an angle of 33.0∘ above the horizontal and is caught by an outfielder 380ft from home plate at the same height from which it left the bat. 1. What was the initial speed of the ball?

More Similar Questions