A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

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A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

eight of h2 = 1.38 m. Neglect air resistance. Randomized Variables m = 0.49 kg h1 = 2.25 m h2 = 1.38 m show answer No Attempt 33% Part (a) Select an expression for the impulse I that the baseball experiences when it bounces off the concrete.

Physics

1 answer:

Yuliya22 [3.3K] · Answer 1 · 2026-04-11T11:37:47+03:00

Response:

Clarification:

Impulse is equal to change in momentum

mv - mu, where v and u represent the final and initial velocities during the surface impact

For the downward motion of the baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

This becomes the initial velocity upon impact.

For the upward movement

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes the final velocity post-impact

change in momentum is

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

=.7056 N.s.

Impulse exerted by the floor in the upward direction is

=.7056 N.s