Question

A batter hits a pitched ball when the center of the ball is 0.9 m above the ground. The ball leaves the bat at an angle of 45 degrees with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 100 m. The ball is hit toward a 7 m high fence that is 88 m away from the launch point. Find the distance between the top of the fence and the center of the ball when the ball reaches the fence

Answer 1

since we are provided with range= 100 m and the launch angle= 45 degrees. From this, we can get the initial speed from the equation:

R=v0^2 sin(2 theta)/g

where v0 is initial speed and g is accleration due to gravity

sin (2x45)=1, so we have

v0^2=gR= 100x9.81=981
v0=31.32m/s

from this, we can find the initial x and y components of speed:

v0x=v0cos45=31.32 cos45=22.14 m /s;

we also know that this is the horizontal speed throughout the trip since there are no forces in the horizontal direction to change the horizontal motion

v0y=v0sin45=31.32 cos45=22.14 m/s (the vertical speed varies because of the vertical force of gravity)

so if the horizontal speed is always 22.14 m/s, then we know that the ball is 88m away when t=88m/22.14m/s =
3.98 s

now, use the equation of motion in the y(t) direction:

y(t)=y0+v0yt-1/2gt^2
y(3.98)=0.9 m + 22.14m/s x 3.98s - 1/2 (9.8m/s)(3.98)^2
y(3.98)=11.4m

so the center of the ball is 4.4 m above the top of the fence