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 0.312 s to go past the window. What was the ball's initial velocity? Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Answer 1

54 An objectin free fall experiences an acceleration gof-9.81m/s*2. (The -signindicates a downwardacceleration) Given Supposev is the velocity at 7.60m offthe ground, and vh is the velocity at9.85moffthe ground Hence, From first equation, Substitutingto second equation 2.00 ,9.81x0.312 0.312 = 6.88m/s The initialvelocity is 6.88 20-9.81)07.5) - 194.48 v-194.48 13.94m/s