44.1 m ઊંચાઇ ધરાવતા એક પુલ ઉપરથી પથ્થરને

Time taken by first stone to reach the water surface from the bridge be $t$, then

$h = ut + \frac{1}{2}g{t^2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8{t^2}$

$t = \sqrt {\frac{{2 \times 44.1}}{{9.8}}} = 3\;sec$

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone

$ = 3 - 1 = 2\;sec$

Hence $44.1 = u \times 2 + \frac{1}{2}9.8{(2)^2}$

$ \Rightarrow 44.1 - 19.6 = 2u \Rightarrow u = 12.25\;m/s$