Question

A rock was thrown horizontally from a 100.0 m high cliff. It strikes the ground 90.0m from the base of the cliff. At what speed was it thrown?

Answer 1

To find the initial speed at which the rock was thrown, we can use the principles of projectile motion. Since the rock was thrown horizontally, its initial vertical velocity component is zero. The only force acting on the rock is gravity, which causes it to accelerate downward.

The horizontal distance traveled by the rock, $d=90.0\text{m}$, is the result of the horizontal velocity $v_x$ and the time of flight $t$ of the rock.

We can calculate the time of flight using the vertical motion of the rock. The vertical distance traveled by the rock, $h=100.0\text{m}$, can be used to find the time of flight. We'll use the equation:

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

Where:

• $h$ is the vertical distance traveled (100.0 m)
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²)
• $t$ is the time of flight (unknown)

Rearranging the equation to solve for $t$:

$t=\sqrt{\frac{2h}{g}}$

Substituting the given values:

t=2⋅100.0 m9.8 m/s2t = sqrt{frac{2 cdot 100.0 , ext{m}}{9.8 , ext{m/s}^2}}t=<span class="m