Question

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball's initial velocity?

Answer 1

Answer:

u =  13.67 m/s

Explanation:

given,

window height = 2 m

window is 7.5 m off the ground on its path up

total distance from the ground to pass the window = 2 + 7.5 = 9.5 m

time taken to go past the window = 1.30 s

using equation of motion

$S = u t +\dfrac{1}{2}gt^2$

$(2+7.5)= u\times 1.3 -0.5\times 9.8\times 1.3^2$

$u\times 1.3 = 9.5 + 8.281$

u =  13.67 m/s

hence, the initial velocity of the  ball is equal to 13.67 m/s