Question

a car of mass 1.032 kg travels towards a cliff. if the car lands 59.1 meters from the base of the cliff and the cliff is 80 m tall, what was the velocity of the car? assume g = 10 m/s^2. round to 1 decimal place. answer

Explanation:

Step1: Calculate time of fall

The car is in free - fall vertically. Using the equation for vertical displacement $y = v_{0y}t+\frac{1}{2}gt^{2}$, where the initial vertical velocity $v_{0y}=0$ m/s, $y = 80$ m and $g = 10$ m/s². So, $y=\frac{1}{2}gt^{2}$, and $t=\sqrt{\frac{2y}{g}}$.
$t=\sqrt{\frac{2\times80}{10}}=\sqrt{16}=4$ s.

Step2: Calculate horizontal velocity

The horizontal displacement $x = 59.1$ m. In the horizontal direction (no acceleration, $a_x = 0$), the equation is $x = v_{x}t$. We know $x$ and $t$, so $v_{x}=\frac{x}{t}$.
$v_{x}=\frac{59.1}{4}=14.8$ m/s.

Answer:

$14.8$ m/s