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2014-10-03T21:58:08+05:30

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So to get the maximum range the angle must be equal to 45°

using the formula for taking out height reached by a projectile

we know

H=\frac{u^{2}sin^{2}{A}}{2g}

so 

R=\frac{u^{2}sin{2A}}{g}

R=\frac{2u^{2}sin{A}cos{A}}{g}

R=\frac{4u^{2}sin{A}cos{A}}{2g}

Putting H

We have

\frac{R}{4}=\frac{u^{2}sin^{2}{A}cos{A}}{2g*sin{A}}

\frac{R}{4}=H*\frac{cos{A}}{sin{A}}

Putting A = 45°

we get 

H=\frac{R}{4}
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ok plz mark so that i could edit
done
thank you
hope it helps
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2014-10-03T23:46:08+05:30

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Time to reach the maximum height H = T = u Sin θ / g

         Range = R = (u Cos θ) * (2 * u sin θ)/g =  u² Sin 2θ / g

Range is maximum when 2θ = 90⁰  and  Sin 2θ = 1

           R_max = R = u² / g

Maximum Height reached = H = u² Sin² θ / 2 g = u² / 4 g = R / 4

===================================

Formula for relation between maximum height and range is :
            H / R  =  tan θ / 4

       So H = R / 4           as   tan 45⁰ = 1

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θ = 45⁰, the projectile travels maximum range for the same initial speed.