A) 8 m
B) 10 m
C) 12 m
D) 16 m
Correct Answer: B
Solution :
Let particle thrown with velocity u and its maximum height is H then \[H=\frac{{{u}^{2}}}{2g}\] When particle is at a height \[H/2\], then its speed is 10m/s From equation \[{{v}^{2}}={{u}^{2}}-2gh\] \[{{(10)}^{2}}={{u}^{2}}-2g\left( \frac{H}{2} \right)={{u}^{2}}-2g\frac{{{u}^{2}}}{4g}\]\[\Rightarrow {{u}^{2}}=200\] Maximum height \[\Rightarrow H=\frac{{{u}^{2}}}{2g}=\frac{200}{2\times 10}=10\ m\]
You need to login to perform this action.
You will be redirected in
3 sec
