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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.`
7 years ago

HRIDAM BASU
16 Points
```										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)
```
7 years ago
Shrey Sharma
24 Points
```										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
```
7 months ago
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