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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.

bharath devasani , 15 Years ago

Grade 12

2 Answers

HRIDAM BASU

Last Activity: 15 Years ago

At the highest point, v=v_x

At half the highest point, v_y²=u²sin²θ -2g(h/2)

v_x=ucosθ , so, v²=(v_x²+v_y²)=u²-gh

By ques, u²cos²θ=(2/5)(u²-gh)

So, θ= cos^-1√(2/5)(1-gh/u²)

Shrey Sharma

Last Activity: 8 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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