Answer
12 m/s
Work Step by Step
Time taken for the first stone to strike the water is obtained as below:
$\Delta y= ut+ \frac{1}{2}at^{2}$
$-43.9\,m= 0\times t+\frac{1}{2}\times(-9.8\,m/s^{2})\times t^{2}$
$\implies t=\sqrt {\frac{-43.9\,m}{-4.9\,m/s^{2}}}=3.0\,s$
t for the second stone is (3.0-1.00)s =2.0 s.
Substituting the value of t in $\Delta y= ut+ \frac{1}{2}at^{2}$, the initial velocity of the stone is obtained.
$-43.9\,m= u\times2.0\,s+\frac{1}{2}\times(-9.8\,m/s^{2})\times (2.0\,s)^{2}$
$\implies u=\frac{-43.9\,m+19.6\,m}{2.0\,s}=-12\,m/s$
Initial speed= 12 m/s
