InterviewSolution
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InterviewSolution
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The velocity of a projectile when at its greatest height is sqrt((2)/(3)) of its velocity when at half of its greatest height find the angle of projection |
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Answer» Solution :STEP 1 : we know that, velocity of a projectile at half of MAXIMUM height `U= sqrt((1+ cos^(2) theta)/(2))` Step 2 :given that `u cos theta sqrt((2)/(5)) xx u sqrt(((1+ cos^(2) theta))/(2))` Squaring on both sides `u^(2) cos^(2) theta=(2)/(5) u^(2)((1+cos^(2) theta)/(2))` `10 cos theta=2+ 2 cos theta` `rArr 8 cos^(2) theta=2 rArr cos^(2) theta =(1)/(4) rArr theta =60^(@)` |
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