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