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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

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

Grade:11

2 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
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 )
so the required answer is:
75^{0}
Thanks & Regards
Sumit Majumdar,
askIITians Faculty
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

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