The speed of a projectile when it is at is greatest height is 25 times its speed at half the maximum height the angle of projection is

VH=25 VH2 Since speed at a point at height ‘h’ is V=u2−2ghSince H=u2sin2θ2g VH=u2−2gh=ucosθ and VH2=U2−2g(H2)VH2=U2−gH ⇒VH2=u2−u2sin2θ2 ⇒VH2=25VH22 u2cos2θ=25(u2−u2sin2θ2) ⇒5u2cos2θ=2u2(1−sin2θ2) ⇒5cos2θ=2−sin2θ 5−5sin2θ=2−sin2θ ⇒3=4sin2θ ∴sin2θ=34 sinθ=32 ⇒θ=600

The speed of a projectile when it is at is greatest height is 25 times its speed at half the maximum height the angle of projection is

VH=25 VH2 Since speed at a point at height ‘h’ is V=u2−2ghSince H=u2sin2θ2g VH=u2−2gh=ucosθ and VH2=U2−2g(H2)VH2=U2−gH ⇒VH2=u2−u2sin2θ2 ⇒VH2=25VH22 u2cos2θ=25(u2−u2sin2θ2) ⇒5u2cos2θ=2u2(1−sin2θ2) ⇒5cos2θ=2−sin2θ 5−5sin2θ=2−sin2θ ⇒3=4sin2θ ∴sin2θ=34 sinθ=32 ⇒θ=600

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The speed of a projectile when it is at is greatest height is $\sqrt{\frac{2}{5}}$ times its speed at half the maximum height the angle of projection is

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$30^{0}$

$\tan^{- 1} (\frac{3}{4})$

$45^{0}$

$60^{0}$

answer is D.

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Detailed Solution

$V_{H} = \sqrt{\frac{2}{5}}$ $V_{\frac{H}{2}}$ Since speed at a point at height ‘h’ is $V = \sqrt{u^{2} - 2 g h}$
Since $H = \frac{u^{2} \sin^{2} θ}{2 g}$
$V_{H} = \sqrt{u^{2} - 2 g h} = u \cos θ$ and $V_{\frac{H}{2}} = \sqrt{U^{2} - 2 g (\frac{H}{2})}$
$V_{\frac{H}{2}} = \sqrt{U^{2} - g H} \Rightarrow V_{\frac{H}{2}} = \sqrt{u^{2} - \frac{u^{2} \sin^{2} θ}{2}} \Rightarrow V_{H}^{2} = \frac{2}{5} V_{\frac{H}{2}}^{2} u^{2} \cos^{2} θ = \frac{2}{5} (u^{2} - \frac{u^{2} \sin^{2} θ}{2}) \Rightarrow 5 u^{2} \cos^{2} θ = 2 u^{2} (1 - \frac{\sin^{2} θ}{2}) \Rightarrow 5 \cos^{2} θ = 2 - \sin^{2} θ 5 - 5 \sin^{2} θ = 2 - \sin^{2} θ \Rightarrow 3 = 4 \sin^{2} θ ∴ \sin^{2} θ = \frac{3}{4} \sin θ = \frac{\sqrt{3}}{2} \Rightarrow θ = 60^{0}$