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A boy standing on a 21.1 m tall cliff throws a rock up in the air with an initial velocity of 29.8 m/s. how much time (in s) will the rock take to reach the water below?

User Datosula
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Final answer:

To find the time for a rock thrown upward to fall back to a lower point, we use kinematic equations with the initial velocity, gravity, and height. A quadratic equation is formed and solved for time, with the positive value representing the answer.

Step-by-step explanation:

To determine the time it takes for a rock thrown upwards from a cliff to reach the water below, we use the kinematic equations for uniformly accelerated motion. Assuming negligible air resistance and acceleration due to gravity (g) as 9.81 m/s2, we apply the following kinematic equation:

h = v0t + (1/2)gt2

where h is the height the rock will fall, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity. Substituting known values, the equation becomes:

-21.1 m = (29.8 m/s)t + (1/2)(-9.81 m/s2)t2. The negative value for h signifies that the rock is moving in the opposite direction of the initial velocity.

This is a quadratic equation in the form of at2 + bt + c = 0, which we can solve for t using the quadratic formula t = (-b ± √(b2 - 4ac))/(2a). Here, a = -4.905 m/s2, b = 29.8 m/s, and c = -21.1 m. After solving, we discard the negative time value and retain the positive one, which is the actual time the rock takes to hit the water.

User Markus Rautopuro
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