(A) Applying energy conservation to find speed just before hitting floor,
m g h1 = m v1^2 / 2
v1 = sqrt(2 x 9.81 x 3.65) = 8.46 m/s downward
after the collision,
v2 = sqrt(2 x 9.81 x 2.3) = 6.72 m/s upward
v1 = - 8.46 m/s and v2 = 6.72 m/s
(A) Impulse = change in momentum = 0.1685(6.72 + 8.46)
= 2.56 kg m/s
(B) impulse = F t
2.56 = F (0.187)
F =13.8 N
(C) change in energy = 0.1685 x 9.81 (3.65 - 2.3)
= 2.23 J
Need help with part a,b, & c Class Management Help h 6 test Begin Date: 10/27/2017...
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