Person A is at the top of a cliff. Person B is at the bottom of the cliff. Person B throws a ball straight up toward Person A. How much speed does it have when it gets to Person B? (more info below)

Question

Person A is at the top of a cliff. Person B is at the bottom of the cliff. Person B throws a ball straight up toward Person A. It leaves her hand with speed = v, and 3.0 seconds later, the ball reaches its highest point – right where Person Acan reaches out and grabs it at the top of the cliff. Then person A throws the ball downward toward Person B. It comes out of his hand at the same speed v that Bonnie threw the ball.

Initially, I tried to set up the avg acceleration equation to solve: (-9.8 m/s^2 = 0 m/s - v)/3 s but I got 29.4 m/s as the final velocity however my textbook states the answer is 41.6 m/s.

Any help would be much appreciated

Answer 1

t_peak = v₀/g. (using = - 9.8, v₀ = 29.4 m/s as you said)

This is the part that you missed.... A doesn't leave the ball alone. He throws it downward (or swats it down) at a velocity of -29.4 m/s. Now, the ball will not take as long to go the distance, so we need to calculate the height of A.

You can find it as v_avgt for B's throw upwards or (29.4+0)/2 * 3 s = 44.1 m

Now you can calculate the v at B from A as v = (-29.4 m/s)² - 2(-9.8 m/s²)(-44.1 m) or v = 41.6 m/s

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