A) 5.20m
B) 4.33m
C) 2.60m
D) 8.66m
Correct Answer: D
Solution :
The ball will be at point \[P\] when it is at a height of \[10\,\,m\] from the ground. So, we have to find distance\[OP\], which can be calculated direct considering it as a projectile on a levelled\[(OX)\]. \[OP=R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[=\frac{{{10}^{2}}\times \sin (2\times {{30}^{o}})}{10}\] \[=\frac{10\sqrt{3}}{2}=5\sqrt{3}\] \[=8.66\,\,m\]You need to login to perform this action.
You will be redirected in
3 sec