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A boy standing on a 46.8 m tall cliff throws a rock up in the air with an initial velocity of 20.2 m/s. how much time (in s) will the rock take to reach the water below?

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Final answer:

The rock will take approximately 3.1 seconds to reach the water below.

Step-by-step explanation:

To find the time the rock takes to reach the water below, we can use the equations of motion. Since the rock is thrown up vertically, it will take the same amount of time to reach the top of its trajectory and come back down, assuming negligible air resistance.

Using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time, we can rearrange the equation to solve for t.

Plugging in the given values, we have 46.8 = (1/2)(9.8)t^2, which simplifies to t^2 = 9.6. Taking the square root of both sides gives us t ≈ 3.1 seconds.

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