# A target is fixed on top of a pole 13 m high. A person standing at a distance 50 m from the pole is capable of projecting a stone with velocity 10*rootg(underroot 9.8 m/s^2). If his aim is to strike the target in least time what should be the angle of elevation with time

Sumit Majumdar IIT Delhi
8 years ago
Solution:
The formulae for range and maximum height are given by:
$R=\frac{u^{2}sin\left ( 2\Theta \right )}{g} H=\frac{u^{2}sin^{2}\Theta }{2g}$
Thus in the given case:
$tan\Theta =\frac{R}{H} \Rightarrow \Theta =tan^{-1}\left ( \frac{R}{H} \right )$
$75^{0}$
Thanks & Regards
Sumit Majumdar,
Ph.D,IIT Delhi
muktesh singh
16 Points
8 years ago
Sir this is not the right answer, answer is tan inverse 3/5. In the question they have asked for the least time