A basketball player is running at a constant speed of 2.5m/s when he tosses a ba

Question

A basketball player is running at a constant speed of 2.5m/s when he tosses a ball upward with a speed of 6.0m/s. How far does the player run before he catches the ball? Ignore air resistance.

Explanation / Answer

Greetings... well u know that initial horizontal velocity is 6 m/s and initial vertical velocity is 2.5 m/s first u need to find the time the ball is in the air, using the vertical velocity and the formula -> change in y=v(initial)t+(1/2)at^2 your change in y is 0, initial velocity is 6, and acceleration is g, or -9.8 m/s so your equation is 0=6t-4.9t^2 solve for t t(6-4.9t)=0 t=1.224s (u don't use 0; because that assumes the ball never went anywhere) to find out how far he runs, u use change in x change in x=v(initial)t +(1/2)at^2 (u can also use distance=rate x time) velocity is 2.5 and acceleration is 0 because hes moving at constant velocity, t is 1.224s x=2.5(1.224) x=3.06m Wish you success!

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