Question

A stone is thrown horizontally from a cliff 100 ft high. The initial velocity is 20 ft. How far from the base of the cliff does the stone strike the ground?

Answer 1

To solve this problem, we can use the equations of motion for projectile motion. Since the stone is thrown horizontally, its initial vertical velocity is 0 ft/s.

We can use the following equation to find the time it takes for the stone to reach the ground:

$h=\frac{1}{2}g{t}^2$

where $h$ is the vertical displacement (negative since the stone moves downward), $g$ is the acceleration due to gravity (approximately 32.2 ft/s²), and $t$ is the time taken.

In this case, the initial vertical displacement ($h$) is -100 ft, and we want to find the time it takes for the stone to reach the ground.

Substituting the values, we have:

-100 ft = (1/2) × 32.2 ft/s² × t²

Simplifying the equation:

-100 ft = 16.1 ft/s² × t²

Dividing both sides by 16.1 ft/s²:

t² = -100 ft / (16.1 ft/s²)

t² ≈ 6.2118 s²

Taking the square root of both sides, we find:

t ≈ 2.492 s

Now that we know the time it takes for the stone to reach the ground, we can find the horizontal distance traveled using the equation:

d = v × t

where $d$ is the horizontal distance traveled (what we want to find), $v$ is the initial horizontal velocity (20 ft/s), and $t$ is the time taken.

Substituting the values, we have:

d = 20 ft/s × 2.492 s

d ≈ 49.84 ft

Therefore, the stone strikes the ground approximately 49.84 ft from the base of the cliff.