For what angle range becomes double of height in a projectile motion?

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For what angle range becomes double of height in a projectile motion?

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Answer 1 · 2018-06-09T02:32:27

${tan}^{- 1} 2$ $tan^-1 2$

Explanation:

The range for projectile is $R = \frac{u^{2}}{g} sin (2 α)$ $R = u^2/g sin(2 alpha)$
and the maximum height is given by $H = \frac{u^{2} {sin}^{2} α}{2 g}$ $H = (u^2sin^2 alpha)/(2g)$

Thus, to get $R = 2 H$ $R = 2H$ we must have

$\frac{u^{2}}{g} sin (2 α) = 2 \frac{u^{2} {sin}^{2} α}{2 g} \Rightarrow$ $u^2/g sin(2 alpha) = 2(u^2sin^2 alpha)/(2g) implies$

$sin (2 α) = {sin}^{2} α \Rightarrow$ $sin(2alpha) = sin^2 alpha implies$

$2 sin α cos α = {sin}^{2} α \Rightarrow$ $2sin alpha cos alpha = sin^2 alpha implies$

$tan α = 2$ $tan alpha = 2$

Thus, the angle of projection must be ${tan}^{- 1} 2$ $tan^-1 2$

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For what angle range becomes double of height in a projectile motion?

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