Question

A car drives straight off the edge of a cliff that is 57m high. The investigator at the scene of the accident notes that the point of impact is 130m from the base of the cliff. How fast was the car travelling when it went over the cliff (in m/s)?
A. 36
B. 37
C. 38
D. 39

Answer 1

As we are dealing with projectile motion here, let us consider the vertical motion and horizontal motion of the car separately. By applying Newton’s equation of motion for vertical motion, we will get the time of flight. Now substituting this along with the given horizontal range found by the investigator will give you the horizontal velocity of the car.

Formula used: Newton’s equation of motion,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$

Complete step by step answer:
In the question we are given a car that is driven straight of the edge of a cliff of 57m height. The point of impact is found to be 130m from the cliff by an investigator. We are asked to find the velocity of the car when it went over the cliff.

We know that, at that point of time when the car went over the cliff the vertical component of velocity is zero and the vertical displacement of the car in the given case will be the height of the cliff.
From Newton’s law of motion we have,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
But for vertical motion we have,
$u=0$
$s=57m$
$\Rightarrow t=\sqrt{\dfrac{2s}{g}}$
$\Rightarrow t=\sqrt{\dfrac{2\times 57}{9.8}}$
$\therefore t=3.41s$
Hence we found the time of flight of the given projectile motion to be 3.41s.
Now, we could consider the horizontal motion,
The horizontal displacement that is, the range of the projectile is found to be 130m by the investigator.
We know that by definition horizontal velocity is given by,
${{v}_{x}}=\dfrac{x}{t}$
Where, x is the horizontal range and t is the time of flight. So,
${{v}_{x}}=\dfrac{130}{3.41}$
$\therefore {{v}_{x}}=38.12m{{s}^{-1}}$
Since the horizontal velocity remains constant throughout a projectile motion, this will be the same velocity when the car went off the cliff. Therefore, we found the velocity of the car when it went over the cliff to be approximately $38m{{s}^{-1}}$.

So, the correct answer is “Option C”.

Note: We have few points that have to be kept in mind while dealing with projectile motion. The horizontal velocity remains constant throughout, so, there will be no horizontal acceleration and the acceleration in the vertical direction will be that due to gravity. Though the initial vertical velocity is zero for a horizontal projectile, it gradually develops a vertical component of velocity.