Question

A person jumps from the roof of a house 4.4-mhigh. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.71 m .

A.) If the mass of his torso (excluding legs) is 45 kg , find his velocity just before his feet strike the ground.

B.) If the mass of his torso (excluding legs) is 45 kg , find the magnitude of the average force exerted on his torso by his legs during deceleration.

Answer 1

Let the downward direction be the +ve y direction
A) Y_O = 0 m, Y = 4.4 m, v₀ = 0 m/s, a = 9.8 m/s², v = ?
v² =u²+ 2a( Y- Y_O )
v = = 9.28 m/s

B) The person has to stop from 9.28 m/s to 0 m/s in0.71 m
v_{0 =} 9.28 m/s , v = 0 m/s, y - y₀ = 0.71 m, a = ?
     v² =u²+ 2a( Y- Y_O )
a = - 9.28²/ (2 *0.71) = -60.64 m/s²
Total force on th e torso = mg - F

Newton's 2nd law ,  mg - F = ma
therefore, F = 45 * 9.8 + 45 *60.64 = 3169.8 N Force acts upward