A stone is dropped into water from a bridge $44.1\,m$ above the water. Another stone is thrown vertically downward $1$ sec later. Both strike the water simultaneously. What was the initial speed of the second stone.......$m/s$
A stone is dropped into water from a bridge $44.1\,m$ above the water. Another stone is thrown vertically downward $1$ sec later. Both strike the water simultaneously. What was the initial speed of the second stone.......$m/s$
$12.25$
$14.75$
$16.23$
$17.15$
A stone is dropped into water from a bridge $44.1\,m$ above the water. Another stone is thrown vertically downward $1$ sec later. Both strike the water simultaneously. What was the initial speed of the second stone.......$m/s$
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$
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