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
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
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 :)
To calculate the height to which the baseball rises, you can use the equations of motion. The key equation we will use is the vertical displacement formula:
Δy = v₀y * t + (1/2) * a * t²
Where:
- Δy is the vertical displacement (height)
- v₀y is the initial vertical velocity (upward)
- t is the time the baseball is in the air
- a is the acceleration due to gravity (g)
In this case, we are given the initial vertical velocity and time. We know that the initial velocity is in the upward direction, so we can assume v₀y = 37.8 m/s. Also, the acceleration due to gravity is approximately -9.8 m/s², as it acts in the downward direction.
Plugging in the given values into the equation, we get:
Δy = (37.8 m/s) * (7.72 s) + (1/2) * (-9.8 m/s²) * (7.72 s)²
Calculating this expression will give us the answer to the height the ball rose:
Δy = 73 m
Therefore, the height to which the ball rises is 73 meters.