Question

Challenge a rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0° above the horizontal. Find its velocity when it hits the ground below?

Answer 1

To find the velocity when the rock hits the ground, we need to consider the vertical and horizontal components of motion separately and then combine them. The vertical component can be calculated using the equation for vertical displacement, while the horizontal component is the same as the initial velocity. Solving for the time it takes for the rock to hit the ground and plugging in the values, we find that the velocity is 7.0 m/s.

Step-by-step explanation:

To find the velocity of the rock when it hits the ground, we need to consider the vertical and horizontal components of the motion separately and then combine them.

First, let's find the time it takes for the rock to hit the ground using the vertical motion. We can use the equation for the vertical displacement:

h(t) = h0 + v0yt - (1/2)gt2

We know that the initial vertical velocity is 7.0 m/s and the acceleration due to gravity is 9.8 m/s2. Assuming the ground is at height 0, we can solve for t:

2.35 = 50 + 7sin(53)t - 4.9t2

This equation is a quadratic equation, so we can solve it using the quadratic formula. Once we have t, we can find the horizontal component of the velocity:

vx = v0x

Since the rock is thrown horizontally, the initial horizontal velocity is the same as the initial velocity, v0 = 7.0 m/s. Therefore, the velocity of the rock when it hits the ground is 7.0 m/s.