A rock is thrown from a cliff with a horizontal velocity of 12.6m/s it lands in the water below 1.45 seconds later, how high is the cliff?

The height of the cliff is 28.57 m.

Explanation:

Given;

initial velocity of the rock, u = 12.6 m/s

time of motion, t = 1.45 s

The height of the cliff is given by;

h = ut + ¹/₂gt²

h = (12.6 x 1.45) + ¹/₂(9.8)(1.45)²

h = 18.27 + 10.30

h = 28.57 m

Therefore, the height of the cliff is 28.57 m.

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