Question

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically

downward with a velocity 20 ms-1

. The second stone will overtake the first after travelling a distance of

Answer 1

Let's imagine that we have axis OY starting from the top of the tower. Then for the first stone coordinate will depend on time as

$\displaystyle y_1= \frac{gt^2}{2}$

For the second stone:

$\displaystyle y_2 = vt + \frac{gt^2}{2}$

We need to take into account that the second stone is thrown 1 second later. This could be done by

$t_2 = t_1-1$ (when t1=1, t2=0 - just started to fall down)

The time, when the second stone meets the first stone is a solution of equation

$y_1=y_2$

$\displaystyle \frac{gt_1^2}{2} = v (t_1-1) + \frac{g (t_1-1)^2}{2}$

$\displaystyle \frac{gt_1^2}{2} = v (t_1-1) + \frac{g (t_1^2-2t_1+1)}{2}$

$\displaystyle 0 = v (t_1-1) + \frac{g (-2t_1+1)}{2} = vt_1 - v - gt_1 +g/2 = t_1 (v-g) -v +g/2$

$\displaystyle t = \frac{v-g/2}{v-g} = \frac{20-4.9}{20-9.8}= \frac{15.1}{10.5} =1.44 \;s$

Let's put this value in any equation y(t) (since at this time y1=y2 it doesn't matter where to put this time):

$\displaystyle y_2 = y_1 = \frac{gt^2}{2} = \frac{9.8 \cdot 1.44^2}{2}= \frac{20.32}{2}=10.16\, m$

Answer: 10.16 m.