At t=0, a stone is dropped from a cliff above a lake; 1.6 s later another stone is thrown downward from the same point with an initial speed of 32 m/s. both stones hit the water at the same instant. find the height of the cliff.

27.52 meters is the height of the cliff.

Explanation:

Second equation of motion:

u = Initial velocity

t = Time taken to cover h  distance

g = Acceleration due to gravity

Let the height of the cliff be h

Stone-1:

Initial velocity of the stone = u = 0 m/s

Time to cover h height of cliff, t = ?, g =

..[1] (Second equation of motion)

Stone-2:

Initial velocity of the stone = u' = 32 m/s,

Time to cover h height of cliff = t' = (t-1.6),

g =

..[2]

Both stones are covering same distance or height of the cliff: [1]= [2]

On solving for t:

t = 2.37 seconds

h = 27.52 m

27.52 meters is the height of the cliff.

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