<span>Let's find out the ball's acceleration
V^2 = U^2 + 2as
Where the distance S = 6 feet and U = 45.
Suppose he just took off so V = 0 and we are left with U^2 =2as
(45)^2 = 2a(6)
a = 2025/12 = 168.75 ft/s^2
Also S = ut + 1/2 at^2
Here S = 5ft and U= 0; a = 168.75. So we have 5 = (1/2) * 168.75 * t^2
So we have 5 * 2 = 168.75 t^2
T^2 = âš 10/ 168.75 = 0.00596 = âš0.006
T = 0.0077s</span>
