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# 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
6 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,