Answer
11.35 m/s - Downward direction
Work Step by Step
Let's take,
The initial velocity of the second stone $=u_{2}$
Displacement of the two stones $=15\space m $
Time in the air for the first stone $=t_{1}$
Time in the air for the first stone until 3.2m displacement$=t_{2}$
Time in the air for the second stone $=t_{1}-t_{2}$
Gravitational Acceleration $=9.8\space m/s^{2}\downarrow$
Let's apply equation 2.8, $S=ut+\frac{1}{2}at^{2}$ to the first stone, until it comes to the ground.
$\downarrow S=ut+\frac{1}{2}at^{2}$
$15\space m=0\space m/s\times t_{1} + \frac{1}{2}\times9.8\space m/s^{2}t_{1}^{2}$
$t_{1}^{2}=\frac{30}{9.8}s^{2}=>t_{1}=1.75\space s$
Let's apply equation 2.8, $S=ut+\frac{1}{2}at^{2}$ to the first stone, until it comes to 3.2 m.
$\downarrow S=ut+\frac{1}{2}at^{2}$
$3.2\space m=0\space m/s\times t_{2}+\frac{1}{2}\times 9.8\space m/s^{2}\times t_{2}^{2}$
$t_{2}^{2}=\frac{6.4}{9.8}s^{2}=>t_{2}=0.808\space s$
$t_{1}-t_{2}=1.75\space s-0.808\space s=0.94\space s$
Let's apply equation 2.8, $S=ut+\frac{1}{2}at^{2}$ to the second stone, until it comes to the ground.
$\downarrow S=ut+\frac{1}{2}at^{2}$
$15\space m=u_{1}(t_{1}-t_{2})+\frac{1}{2}\times 9.8\space m/s^{2}(t_{1}-t_{2})^{2}$
$15\space m=u_{1}(0.94\space s)+\frac{1}{2}\times 9.8\space m/s^{2}(0.94\space s)^{2}$
$u_{1}=\frac{15-4.33}{0.94}m/s=11.35\space m/s$
