Answer:
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
or .5 (one-half)