A ball is thrown straight up it passes a 2.00 m high window 7.5 meters off the ground on its path up it takes 1.3 s to go past the window. what was the balls initial velocity?

Question

Tyrell Kramer

Physics

26 July, 07:40

A ball is thrown straight up it passes a 2.00 m high window 7.5 meters off the ground on its path up it takes 1.3 s to go past the window. what was the balls initial velocity?

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Jaquan · Answer 1

Let's say the velocity at the bottom of the window was "v."

s = v*t + ½at²

2 m = v * 1.3s - 4.9m/s² * (1.3s) ² = v * 1.3s - 8.3 m

v = 10.3m / 1.3s = 7.9 m/s

Then the initial speed was

V = √ (v² + 2as) = √ (7.9m/s² + 2 * 9.8m/s² * 7.5m) = 14 m/s ◄ initial velocity

(after rounding to 2 digits from 14.5 m/s).