A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 21m/s.The cliff is 48m above the waters surface.How long does it take for the stone to fall in the water? g = 9.81 m/s^2.
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When the person kicks the stone over the edge of the cliff, the stone has a horizontal velocity of 21 m/s. The vertical velocity is 0.
The cliff is the 48 m above the water surface. We need to determine the time taken by the stone to fall in the water. This is the time require for a body with velocity equal to 0 but which is being accelerated at 9.8 m/s^2 to travel a distance of 48 m.
Use the relation D = u*t + (1/2)*a*t^2. Here u or the initial velocity is 0. The distance D = 48 and a is 9.8 m/s^2
This gives 48 = 0 + (1/2)*9.8*t^2
=> t^2 = 48*2/9.81
=> t = sqrt(96/9.81)
=> t = 3.12 seconds
The stone falls into the water 3.12 seconds after it is kicked from the cliff.
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