A stone was thrown upwards with avelocity of 56.6m/s from a cliff.If the height of the cliff is 242m, what is the velocity of the stone when it hits the ground?Assume that air resistance is negligible.
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Explanation:
The initial velocity of the stone thrown from the cliff is 5 m/s
The stone strikes the pond near the base after 4 sec.
The stone moves in upward direction initially and then after reaching a maximum height, it starts to come down and reaches the base of the cliff.
In overall motion of the stone, the net displacement in the opposite direction of the initial velocity.
S=ut−21at2
Substitute the values in above expression:
S=(5×4)−21×10×42
S=20−80=−60 m
So, the height of the cliff is 60 m.
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