Atennis ball is thrown off a cliff at an angle of 26â° above the horizontal. the tennis ball hits the ground below the cliff 3.2 seconds later at a horizontal distance of 200 ft from where it was thrown. how high is the cliff?

h = 56.57 ft

Explanation:

given,

angle at which tennis ball was hit = 26°

time taken to hit the cliff = 3.2 s

horizontal distance = 200 ft.

calculation of velocity

horizontal velocity,v_x = v cos θ

distance = speed x time

 200 = v cos 26° x 3.2

  v = 69.54 ft/s

now, calculation of time to reach maximum height

final velocity = 0 ft/s

vertical velocity = v sin θ

using equation of motion

v = u + at

0 = v sin 29° - g t

32.2 t = 69.54 sin 29°

t = 1.053 s

vertical distance of maximum height travel by the ball

 x = 17.64 ft

now,

height of the cliff

distance traveled by the ball from top point to ground= x + h

time taken = t' = 3.2 - 1.053 = 2.147 s

initial velocity = 0 ft/s

using equation of motion

h = 56.57 ft

height of the cliff is equal to 56.57 ft

the answer is to check all of them! they all apply well at least i think so!