Question

1. you throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

a. how long does it take the stone to reach the bottom of the cliff?
b. how far from the base of the cliff does the stone hit the ground?
c. what are the horizontal and vertical components of the stones velocity just before it hits the ground?

Explanation:

Step1: Analyze vertical - motion to find time

The vertical - motion of the stone is a free - fall motion. The height of the cliff $h = 78.4$ m, and the initial vertical velocity $v_{0y}=0$ m/s. The equation for vertical displacement is $h = v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0$ m/s, we have $h=\frac{1}{2}gt^{2}$, where $g = 9.8$ m/s². Solving for $t$:
\[t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times78.4}{9.8}}=\sqrt{16}=4\ s\]

Step2: Analyze horizontal - motion to find horizontal distance

The horizontal motion of the stone is a uniform - motion with a constant horizontal velocity $v_{x}=5.0$ m/s. The horizontal displacement $x$ is given by the formula $x = v_{x}t$. Substituting $v_{x}=5.0$ m/s and $t = 4$ s, we get $x=5\times4 = 20$ m.

Step3: Analyze vertical - motion to find vertical velocity

The vertical velocity $v_{y}$ at the bottom of the cliff is given by the formula $v_{y}=v_{0y}+gt$. Since $v_{0y}=0$ m/s and $t = 4$ s, $g = 9.8$ m/s², we have $v_{y}=9.8\times4=39.2$ m/s, and $v_{x}=5.0$ m/s.

a.

The time it takes for the stone to reach the bottom of the cliff:
\[t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times78.4}{9.8}} = 4\ s\]

b.

The horizontal distance from the base of the cliff where the stone hits the ground:
\[x=v_{x}t=5\times4 = 20\ m\]

c.

The horizontal component of the velocity $v_{x}=5.0$ m/s (constant in projectile motion without air - resistance), and the vertical component of the velocity $v_{y}=v_{0y}+gt=0 + 9.8\times4=39.2$ m/s.

Answer:

a. The time is $4$ s.
b. The horizontal distance is $20$ m.
c. The horizontal component of velocity is $5.0$ m/s and the vertical component of velocity is $39.2$ m/s.