A ball is launched from 8 yards off the ground and travels in parabolic motion, landing in a net 80 yards away and also 8 yards off the ground. A wall lies 75% of the way down the path, which the ball cleared by 13 yards.

If the ball's maximum height during its flight was 80 yards, how many yards tall is the wall?

The answer to this question is the wall is 45.95 yards tall

Step-by-step explanation:

To solve this, we list out the given variables and the unknowns thus

Height of ball at launch = 8 yards

Distance of net from the ball = 80 yards

Distance of the wall down the path = 75%

Maximum height of the ball = 80 yards

equation of Motion of the ball = parabolic motion =

v² = u² - 2gS

S = 80 - 8 = 72 yards

at maximum height v = 0 thus u² = 2*9.81*72 = 1412.64

u = 37.59 m/s

also v = u - gt and again at max height v = 0

Therefore 37.59 = 9.81*t or t = 3.83 s

If the motion of the ball is free of obstruction then time of flight before the ball just reaches the 8 yards off the ground = 3.83*2 = 7.66 seconds

Taking the initial velocity as zero at maximum height and from the equation

S = ut + 0.5*gt² we get, where S is the heigt of the ball from touching the actual field ground which is 80 yards we have

80 = 0.5*9.81*t²

so that t² = 2*80:9.81 = 16.31 or t = 4.04s

Therefore the total time of flight = 4.04 + 3.83 = 7.87 seconds

if the ball is considered as having a constant horizontal velocity, therefore

at 75% of the way the time it took will be 0.75*7.87 = 5.9 seconds

However time it took the ball to reach maximum height and then starts descent = 3.83s, and the time at which the ball is directly over the wall = 2.07 seconds on the second half just after reaching mximum height

Thus at 2.07 seconds the distance trvelled from the maximum height is

S = ut + 0.5gt² as before where u = 0

hence S = 0.5*9.81*2.07² = 21.05 yards or (80 - 21.05) yards off the ground = 58.95 yards

As stated in the question, the ball cleared the wall by 13 yards therefore the height of the wall is 58.95 - 13 = 45.95 yards