Question

A rock is thrown horizontally with speed v from the top of a cliff of height H.

a.) Identify an equation that can be used to solve for the time it takes the rock to hit the ground.
b.) Rearrange the equation you wrote to solve for the time it will take the rock to hit the water.
c.) Identify an equation that can be used to solve for the horizontal distance between the bottom of the cliff and the place where the rock lands.
d.) Rearrange the equation you wrote to solve for the horizontal distance D between the bottom of the cliff and the place where the rock lands. Answer in terms of H, v, and physical constants if necessary.

Answer 1

For a rock thrown horizontally from a cliff, the time to hit the ground is found using the equation for vertical motion, H = 0.5gt^2, solving for t gives t = sqrt(2H/g).

Step-by-step explanation:

To determine the time it takes for a rock thrown horizontally from a cliff to hit the ground, we can use the equation for vertical motion, since the horizontal and vertical motions are independent:

s = ut + \frac{1}{2}at^2

Where:

• s is the vertical displacement (in this case, the height H of the cliff)
• u is the initial vertical velocity (0 m/s for horizontal throw)
• a is the acceleration due to gravity (9.81 m/s^2)
• t is the time in seconds

For the first part of the question (a and b), since the initial vertical velocity u is zero, the equation simplifies to:

H = \frac{1}{2}gt^2 ⇒ t = \sqrt{\frac{2H}{g}}

For the second part (c and d), to find the horizontal distance D, we use the constant horizontal velocity v and the time t:

D = vt

To solve for D:

D = v\sqrt{\frac{2H}{g}}