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Q.

A ball is projected from the ground with a speed 15 ms−1 at an angle θ  with horizontal so  that its range and maximum height are equal, then tanθ  will be equal to

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answer is 4.

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Detailed Solution

Projection speed of ball,  u=15ms−1
Given that, R = H
 ⇒u2sin2θg=u2sin2θ2g ⇒sin2θ1=sin2θ2 ⇒2sinθ.cosθ=sin2θ2 [∵sin2θ=2sinθ.cosθ] ⇒4=sinθcosθ⇒4=tanθ ⇒tanθ=4
 
 
 
 
 
 

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