A) 8 m
B) 10 m
C) 15 m
D) 20 m
Correct Answer: B
Solution :
Let \[u\] be the initial velocity and H the maximum height attained. |
When at height \[h=\frac{H}{2},\,\] we have |
\[v={{v}_{1}}=10\text{ }m/s\] |
From third equation of motion |
\[v_{1}^{2}={{u}^{2}}-2gh\] |
or \[{{(10)}^{2}}={{u}^{2}}-2g\frac{H}{2}\] ..(i) |
At height H, \[{{v}_{2}}=0\] |
\[v_{2}^{2}={{u}^{2}}-2gH\] |
or \[0={{u}^{2}}-2gH\] ..(ii) |
Subtract Eq. (ii) From Eq. (i), we get |
\[{{(10)}^{2}}=2g\frac{H}{2}\] |
or \[H=\frac{{{(10)}^{2}}}{g}\] |
or \[H=\frac{{{(10)}^{2}}}{10}=10\,m\] |
Alternative: maximum height attained by the stone |
\[H=\frac{{{u}^{2}}}{2g}\] |
When \[h=\frac{H}{2},\,u=10\,m/s\] |
\[\frac{H}{2}=\frac{{{(10)}^{2}}}{2g}\] |
or \[H=\frac{100}{10}=10\,m\] |
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