A student (m = 63 kg) falls freely from rest and strikes the ground. During the

Question

A student (m = 63 kg) falls freely from rest and strikes the ground. During the collision with the ground he comes to rest in a time of 0.040 s. The average force on the ground is +18,000 N. (Upward is + direction). From what height did the student fall? (Neglect air resistance).

Explanation / Answer

Impulse = F(average) * Time. The impulse is the change in momentum. Initial momentum = 63v where v was his final velocity from the free-fall. Final momentum = 0 as he comes to rest. Change in momentum = 63v - 0 = 63v. Now 63v = 18000 * 0.04 => 63v = 720 => v = 720/63 = 11.42 m/s. Now let's apply an equation of motion to find out the height he fell from. v² = u² + 2gh where u = 0 as he started from rest. 130.41 = 0 + 2 * 9.8 * h => h = 130.41/20 = 6.52 metres.

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