If a projectile is projected at angle thetaθ of horizontal and it just passed by touching the tip of two walls of height aa,seperated by a distance 2a2a,then show that range of its motion will be 2a cot(theta/2)2acot(θ2) ?

1 Answer
Mar 16, 2018

Here the situation is shown below,

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So,let after time tt of its motion,it will reach height aa,so considering vertical motion,

we can say,

a=(u sin theta)t -1/2 g t^2a=(usinθ)t12gt2 (uu is the projection velocity of projectile)

Solving this we get,

t=(2u sin theta_-^+sqrt(4u^2 sin^2 theta -8ga))/(2g)t=2usinθ+4u2sin2θ8ga2g

So,one value (smaller one) of t=tt=t (let) is suggesting the time to reach aa while going up and the other (larger one) t=t' (let) while coming down.

So,we can say in this time interval the projectilw horizontally travelled distance 2a,

So,we can write, 2a=u cos theta(t'-t)

Putting the values and arranging,we get,

u^4 sin^2 2theta -8gau^2 cos^2 theta-4a^2g^2=0

Solving for u^2,we get,

u^2 =(8gacos^2 theta _-^+sqrt(64g^2a^2 cos^4 theta+16a^2g^2sin^2 2theta))/(2 sin^2 2theta)

Putting back sin 2theta=2 sin theta cos theta we get,

u^2=(8gacos^2 theta _-^+sqrt(64g^2a^2 cos^4 theta+64a^2g^2sin^2 theta cos^2 theta))/(2 sin^2 2theta)

or, u^2=(8ga cos^2 theta+sqrt(64g^2a^2cos^2theta(cos^2 theta+sin^2 theta)))/(2sin^2 2theta)=(8gacos^2theta+8ag cos theta)/(2 sin^2 2theta)=(8agcostheta(cos theta+1))/(2 sin^2 2theta)

now,formula for range of projectile motion is R=(u^2 sin 2 theta)/g

So,multiplying the obtained value of u^2 with (sin2 theta)/g,we get,

R=(2a(cos theta+1))/sin theta=(2a* 2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))=2a cot (theta/2)