They should look for the car at a distance of 184.8 m from the base of the cliff.
Explanation:
The motion of the car is a projectile motion (acted upon gravity only), therefore it consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity, [tex]g=9.8 m/s^2[/tex]) in the downward direction
First we consider the vertical motion, in order to find the time of flight of the car. We use the following suvat equation:
[tex]s=u t+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction, we have:
s = 95 m is the vertical displacement of the car (the height of the cliff)
[tex]u=0[/tex] is the initial vertical velocity, which is zero since the car is travelling horizontally
t is the time of flight
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find :
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(95)}{9.8}}=4.40 s[/tex]
So the car takes 4.40 s to fall to the ground.
Now we analyze the horizontal motion: the car moves horizontally with a constant velocity of
[tex]v_x = 42 m/s[/tex]
So, the distance covered in a time t is given by
[tex]d=v_x t[/tex]
and by substituting t = 4.40 s, we find the total horizontal distance covered by the car:
[tex]d=(42)(4.40)=184.8 m[/tex]
So, they should look for the car 184.8 m from the base of the cliff.
Learn more about projectile motion:
brainly.com/question/8751410
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