Question

4) A car traveling at 10 m/sec rolls off the edge of a cliff. The cliff is 37.6 m above the ocean. How far from the base
of the cliff does the car strike the water?

Answer 1

The car strikes the water 27.7 m far from the base of the cliff

Explanation:

The motion of the car is a projectile motion. it follows a parabolic path which consists of two independent motions:  

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

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction.

First, we consider the vertical motion. Since this is a free fall motion, we can use the following suvat equation to find the time of flight of the car:

$s=ut+\frac{1}{2}at^2$  

where:

s = 37.6 m is the vertical displacement (the height of the cliff)

u = 0 is the initial vertical velocity of the car

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity  (choosing downward as positive direction)

Solving for t, we find:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(37.6)}{9.8}}=2.77 s$  

Now we use the horizontal motion. This is a uniform motion at constant speed, so the horizontal distance travelled by the car is given by:

$d=v_x t$  

where:

$v_x = 10 m/s$ is the initial horizontal velocity (which is constant)

t = 2.77 s is the time of flight  

Solving for d:

$d=(10)(2.77)=27.7 m$

So, the car strikes the water 27.7 m far from the base of the cliff.

Learn more about projectile motion here:

brainly.com/question/8751410

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