A woman stands on the edge of a cliff and throws a stone vertically downward with an initial speed of 15 m/s. The instant before the stone hits the ground below, it has 2000 J of kinetic energy. If she were to throw the stone horizontally outward from the cliff with the same initial speed of 15 m/s, how much kinetic energy would it have just before it hits the ground?
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According to law of conservation of enregy, we have
K.E. while hitting = P.E. at top + K.E. at top
Since both the stones will have same P.E. at top and K.E. at top(as velocity and heights are the same), they will have the same K.E. while hitting. So, K.E. of second stone = 2000 J
K.E. while hitting = P.E. at top + K.E. at top
Since both the stones will have same P.E. at top and K.E. at top(as velocity and heights are the same), they will have the same K.E. while hitting. So, K.E. of second stone = 2000 J
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