A ball of mass 1.5 kg is thrown upward. It leaves the thrower\'shand with a velo

Question

A ball of mass 1.5 kg is thrown upward. It leaves the thrower'shand with a velocity of 10 m/s. The following questions refer tothe motion after the ball leaves the thrower's hand. Assume thatthe upward direction is positive.

1) How long does it take for the ball to return to the thrower'shand?
2) What is the final velocity of the ball just before it reachesthe hand?
3) What is the change in momentum of the ball?
4) What is the impulse calculated from the change in momentum?
5) What is the average force acting on the ball?

Explanation / Answer

m = 1.5 kg v0 = 10 m/s g = 9.81 m/s2 (1) y = 0 = v0t - (1/2)g t2 Solve for t. (2) v = a t v - v0 = -g t Solve for v. The easy way to solve this is to recognize thatthe ball will have the same speed (a scalar) down that it hadup. Hence, the velocity (a vector) will be -10 m/s. (3) p = mv = m(v - v0) Remember, momentum and velocity are vectors; your answer should benegative. (4) J = p Impulse is defined as the change in momentum, so this answer is thesame as for (3). (5) J = Ft Impulse is also defined as a force acting during a period of time,t (which is your answer from (1)). So, F = J/t You could substitute in your numbers, etc., but you'll get the sameanswer as you would by just recognizing that once the ball is freefrom the hand, the only force acting on it during the time it's inthe air is the force of gravity (the ball's weight), actingdownward: F = -mg

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