Question

A stone is dropped from the top of a cliff and one second later a second stone is thrown vertically upward with a velocity of 20m/s. how far below the top of the cliff will the second stone overtake the first?

Answer 1

Let's take the upwards as the positive direction. Then, we can write the vertical displacement of the first stone dropped from the top of the cliff:

$$y_1=-\dfrac{1}{2}gt^2.$$

Similarly, we can write the vertical displacement of the second stone thrown vertically upward one second later:

$$y_2=v_0(t-1)-\dfrac{1}{2}g(t-1)^2.$$

At the moment when the second stone overtake the first $y_1=y_2$ and we can write:

$$-\dfrac{1}{2}gt^2=v_0(t-1)-\dfrac{1}{2}g(t-1)^2,$$ $$-\dfrac{1}{2}gt^2=v_0t-v_0-\dfrac{1}{2}g(t^2-2t+1),$$ $$-\dfrac{1}{2}gt^2=v_0t-v_0-\dfrac{1}{2}gt^2+gt-\dfrac{1}{2}g,$$ $$20t-20+9.8t-4.9=0,$$ $$t=\dfrac{24.9}{29.8}=0.83\ s.$$

Finally, we can find how far below the top of the cliff will the second stone overtake the first:

$$y_1=-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(0.83\ s)^2=-3.4\ m.$$

The second stone overtake the first 3.4 m below the top of the cliff.