Question

A soccer ball is kicked At 8m/s. It lands on the ground after being in the air for .85 seconds. At what angle is it kicked?

Answer 1

If ball remains in air for total time T = 0.85 s

this is also known as time of flight

In order to find the time of flight we can use kinematics

$\Delta Y = v_y*t + \frac{1}{2}at^2$

so for complete motion its displacement in y direction will be zero

$0 = v_y* 0.85 + \frac{1}{2}(-9.8)(0.85^2)$

$0 = v_y*0.85 - 3.54$

$v_y = 4.165 m/s$

now we know that net velocity of the ball is 8 m/s

while is y direction component we got is vy = 4.165 m/s

now by component method we can say

$v_y = v sin\theta$

$4.165 = 8 sin\theta$

$\theta = sin^{-1}\frac{4.165}{8}$

$\theta = 31.4^0$

so it is projected at an angle of 31.4 degree above horizontal