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Grade 12Mechanics

The velocity of the projectile whenit is at the greatest height is root over 2/5 times its velocity when it is at half of its greatest height. Determine its angle of projection.

Profile image of bharath devasani
16 Years agoGrade 12
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2 Answers

Profile image of HRIDAM BASU
16 Years ago

At the highest point,  v=vx

At half the highest point,  vy2=u2sin2θ -2g(h/2)

vx=ucosθ  ,  so, v2=(vx2+vy2)=u2-gh

By ques,   u2cos2θ=(2/5)(u2-gh)

So, θ= cos-1√(2/5)(1-gh/u2)   

Profile image of Shrey Sharma
9 Years ago
We know that time taken by any body to fall equal heights is in ratio 1:(sqrt2-1)Using this;v= y component of velocity at half the height v=u sin(thetha) -{gt(sqrt2-1)/sqrt2}and by calculating from v^2-u^2= 2as;gt=u sin(thetha);and v^2+u^2cos^2(thetha)=(5/2)u cos(thetha)Therefore thetha is equal to 60

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