A ball of mass 1.5kg is thrown upward. It leaves thethrower\'s hand with a veloc

Question

A ball of mass 1.5kg is thrown upward. It leaves thethrower's hand with a velocity of 10 m/s. The followingquestions refer to the motion after the ball leaves the thrower'shand. Assume that upward is positive. A) How long does it take for the ball to retun to thethrower's hand? B) What's the final velocity of the ball just before itreaches the hand? C) What's the change in momentum of the ball? A ball of mass 1.5kg is thrown upward. It leaves thethrower's hand with a velocity of 10 m/s. The followingquestions refer to the motion after the ball leaves the thrower'shand. Assume that upward is positive. A) How long does it take for the ball to retun to thethrower's hand? B) What's the final velocity of the ball just before itreaches the hand? C) What's the change in momentum of the ball?

Explanation / Answer

The height that the ball travels can be determined byconservation of energy: 1/2mv^2 = mgh -- thus v is: h = v^2/g   ----> mass cancels h = (0.5)(10^2)/9.8 h = 5.1 m -- At maximum height, velocity is zero, so time to reach thethrower's hand is: v(f) = v(i) +a*t t = -10 /-9.8 t = 1.02 s ---> this is the time to reach maximumheight, so for round trip, it is double t = 1.02*2 = 2.0 s -- Momentum is the same, because mass does not change and thevelocity at which the ball is thrown upward with, is thesame velocity at the time it is caught, so: -- m*v(1) = m*v(2) -- Hope this helps.

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