Question

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How long does it take to reach the ground? How far does the rock land from the base of the cliff?

Answer 1

Answer:

t = 4.08 s

R = 40.8 m

Explanation:

The question is asking us to solve for the time of flight and the range of the rock.

Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:

• Δx = v_0 t + 1/2at²

We have these known variables:

• (v_0)_y = 0 m/s
• a_y = -9.8 m/s²
• Δx_y = -20 m

And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.

• -20 = 0t + 1/2(-9.8)t²
• -20 = 1/2(-9.8)t²
• -20 = -4.9t²
• t = 4.08 sec

The time it takes for the rock to reach the ground is 4.08 seconds.

Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.

List out known variables:

• v_0 = 10 m/s
• t = 4.08 s
• a_x = 0 m/s

We are trying to solve for:

• Δx_x = ?

By using the same equation, we can plug these known values into it and solve for Δx.

• Δx = 10 * 4.08 + 1/2(0)(4.08)²
• Δx = 10 * 4.08
• Δx = 40.8 m

The rock lands 40.8 m from the base of the cliff.