Question

At t = 0, a stone is dropped from a cliff above a lake; 2.4 seconds later another stone is thrown downward from the same point with an initial speed of 47 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

Answer 1

Let $T_0$ be the time when 1st stone was thrown. So 2nd stone was thrown at time $T_0 +2.4$s. Let $T_f$ be the time at which they both hit the lake.

The general equation for both stones is $-H=v_it_i-\frac{1}{2}gt_i^2,\quad t_i=1,2$ , H=height

For stone 1, $-H=-\frac{1}{2}gt_1^2$ .......1

For stone 2, $-H=47t_2-\frac{1}{2}gt_2^2$ ......2

and $t_2=t_1-2.4$......3

So using 1, 2 and 3 we get

$\sqrt{\frac{2H}{g}}=\frac{47\times2.4-\frac{1}{2}\times9.8\times2.4^2}{47+9.8\times2.4}$

So H=7.05m