Answer
(a) The slower stone returns to the ground in 3.3 seconds.
(b) The faster stone reaches a maximum height of 9H.
Work Step by Step
Let $v_0$ be the initial speed of the slower stone. Then $3v_0$ is the initial speed of the faster stone.
(a) faster stone:
Let $t_f$ be the total time in the air.
$t_f = 2\times \frac{3v_0}{g}$
slower stone:
$t_s = 2\times \frac{v_0}{g} = \frac{t_f}{3}$
Since $t_f = 10~s$, then $t_s = \frac{10~s}{3} = 3.3~s$.
The slower stone returns to the ground in 3.3 seconds.
(b) slower stone:
$H = \frac{v_0^2}{2g}$
faster stone:
$H_f = \frac{(3v_0)^2}{2g} = 9\times H$
The faster stone reaches a maximum height of 9H.
