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 = v1 = 10 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 \[H=\frac{100}{10}=10\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
