Question

A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 31 m high. When it hits the ground at the base of the cliff, the rock has a speed of 30 m/s.

a. Assuming that air resistance can be ignored, what is the initial speed of the rock?
b. What is the greatest height of the rock as measured from the base of the cliff?

Answer 1

The initial speed of the rock is approximately 16.03 m/s and the greatest height reached by the rock is approximately 13.10 m.

Step-by-step explanation:

To calculate the initial speed of the rock, we can use the equation of motion for vertical motion. The equation is given by:

v^2 = u^2 + 2as

Where v is the final speed (30 m/s), u is the initial speed (what we need to find), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (31 m).

Plugging in the values, we get:
(30 m/s)^2 = u^2 + 2(-9.8 m/s^2)(31 m)

Solving for u, we find that the initial speed of the rock is approximately 16.03 m/s.

The greatest height of the rock can be found using the equation:

v^2 = u^2 + 2as

Here, v is the final speed at the highest point (0 m/s), u is the initial speed (16.03 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (unknown).

Plugging in the values and solving for s gives us:

(0 m/s)^2 = (16.03 m/s)^2 + 2(-9.8 m/s^2)s

By rearranging the equation, we find that the displacement s is approximately 13.10 m.