Question

A distracted driver is driving towards a turn where the edge of the road leads into a 75.0 m cliff. The velocity of the vehicle is 80.0 km/h. The distracted driver doesn’t pay attention to the turn and drives through it and launches the car off the cliff. Find how far from the base of the cliff the vehicle lands?
*Can you please show your work as well as I would like to understand how you got to your answer. Thanks!

Answer 1

As long as the car is on the road, it moves with a constant speed of 80km/h.

As soon as the car starts to fall down the cliff, it follows a parabolic motion. It means that it still moves with constant speed along the x axis, but it also starts to move along the y axis, with constant acceleration (i.e. the acceleration due to gravity).

The good thing about parabolic motions is that the two motions along the x and y axes are completely separable.

So, first of all, we need to know how long it takes for an object to fall for 75m. The equation of a constantly accelerated motion is

$s=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s_0$ is the initial position, $v_0$ is the initial speed, and $a$ is the constant rate of acceleration. In our case, we start from an initial height of 75m, an initial (vertical!) speed of zero, and our acceleration is $-g$. So, our equation becomes

$s=75-\dfrac{g}{2}t^2$

And we want to solve for the time when $s=0$ (i.e. we want to know how long will it take for the object to reach the ground). We have

$0=75-\dfrac{g}{2}t^2 \iff 75=\dfrac{g}{2}t^2 \iff \dfrac{2\cdot75}{g}=t^2 \iff t=\sqrt{\dfrac{150}{g}}$

(I'm discarding the negative solution because it wouldn't make sense)

Now that we've used the vertical motion to find out the falling time, we can go back to the horizontal motion. We know that the car moves for a certain amount of time at a certain speed. So, we simply have to plug our values in the $s=vt$ equation, to get

$s=80\sqrt\dfrac{150}{g}}$

This is how far from the base of the cliff the vehicle lands.