A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by (given g=9.8 m/s²)

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(3) 2.39 kg m s⁻¹. Idea: the ball arrives going down and leaves going up, so the two velocities have opposite signs and their magnitudes add in the change of momentum. Speed on arrival: v₁=√2gh₁=√2(9.8)(10)=14 m/s downwards. Speed on leaving: v₂=√2gh₂=√2(9.8)(5)=9.9 m/s upwards. J=m(v₂-(-v₁))=m(v₂+v₁). =0.1×(9.9+14)=0.1×23.9=2.39 kg m s⁻¹.

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by (given g=9.8 m/s²)

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