A girl of mass m_1 = 60 kilograms springs from a trampoline with an initial upwa

Question

A girl of mass m_1 = 60 kilograms springs from a trampoline with an initial upward velocity of v_i = 8.0 meters per second. At height h = 2.0 meters above the trampoline, the girl grabs a box of mass m_2 = 15 kilograms. (Figure 1)
For this problem, use g = 9.8 meters per second per second for the magnitude of the acceleration due to gravity.

Part A:
What is the speed v_before of the girl immediately before she grabs the box?

Part B:
What is the speed v_after of the girl immediately after she grabs the box?

Explanation / Answer

1) By v^2 = u^2 -2gs =>v^2 = (8)^2 - 2 x 9.8 x 2 =>v = sqrt24.8 = 4.98 m/s 2) By the law of momentum conservation:- =>m1u1+m2u2 = (m1+m2) x v =>60 x 4.98 + 0 = (60+15) x v =>v = 3.98 m/s 3)By v^2 = u^2 - 2gs =>0 = (3.98)^2 - 2 x 9.8 x s =>s = 0.81m =>Total height with respect to the top of the trampoline = 2 + 0.81 = 2.81m

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