Question

Cars driving 15.7 m/s on a flat ground drives off of 25.3 m High cliff how far from the base of the cliff does it land

Answer 1

The car lands 35.6 m from the base of the cliff

Explanation:

The motion of the car is a projectile motion, which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by analyzing the vertical motion, in order to find the time of flight of the car. We do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 25.3 m is the vertical displacement of the car (the height of the cliff)

$u_y=0$ is the initial vertical velocity of the car (because it is moving horizontally)

t is the time of the fall

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find :

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(25.3)}{9.8}}=2.27 s$

Now we can analyze the horizontal motion of the car, which moves with constant horizontal velocity of

$v_x = 15.7 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

and by substituting t = 2.27 s, we find the how far from the base of the cliff the car lands:

$d=(15.7)(2.27)=35.6 m$

Learn more about projectile motion:

brainly.com/question/8751410

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