Q1. A stone of mass 80 grams is released from rest at the top of a vertical cliff. After falling for 3 seconds, it reaches the ground and penetrates 9 cm into the ground.
1. the height of the cliff
2. the average force resisting the penetration
(assume gravitational acceleration g=10m/sec2)
Solution. Assume that the height of the cliff is H(m)and the average force resisting the penetration is F (N). Denote the mass of the stone by m, so m=0.08kg. ...
With good explanations and calculations, the problem is solved.