A missile is launched from the ground making $45^{0}$ with the horizontal to hit a target at a horizontal distance of $300 k m$ . If it is required to hit a target at a horizontal distance of $675 k m$ launched at the same angle with horizontal, find the percentage change in its velocity of projection.

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Solution

$θ = 45 °$
$R = \frac{u^{2} sin 2 θ}{g}$
$R = \frac{u^{2}}{g}$
$u_{300} = \sqrt{300 \times g} = 54.77$
Similarly, $u_{675} = \sqrt{675 g} = 82.15$
% $u = \frac{82.15 - 54.77}{54.77} \times 100 = 47.39$ %

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A missile is launched from the ground making 450 with the horizontal to hit a target at a horizontal distance of 300 km. If it is required to hit a target at a horizontal distance of 675 km launched at the same angle with horizontal, find the percentage change in its velocity of projection.

θ=45°R=u2sin2θgR=u2gu300=√300×g=54.77Similarly, u675=√675g=82.15% u=82.15−54.7754.77×100=47.39%

$θ = 45 °$
$R = \frac{u^{2} sin 2 θ}{g}$
$R = \frac{u^{2}}{g}$
$u_{300} = \sqrt{300 \times g} = 54.77$
Similarly, $u_{675} = \sqrt{675 g} = 82.15$
% $u = \frac{82.15 - 54.77}{54.77} \times 100 = 47.39$ %