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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.

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.

Grade:12

2 Answers

HRIDAM BASU
16 Points
12 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)   

Shrey Sharma
24 Points
5 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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