Question

Q3. Shows that the projectile angle $\theta_0=\tan^{-1}\bigg\lgroup\sf{\dfrac{4h_m}{R}}\bigg\rgroup$, where the symbols have their usual meaning.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let first represent what symbol means,

$\rm :\longmapsto\:h_m \: = \: maximum \: height \: of \: projectile$

$\rm :\longmapsto\:R \: = \: horizontal \: range \: of \: projectile$

$\rm :\longmapsto\:u_0 \: is \: the \: initial \: velocity$

We know,

The maximum height of projectile is given by,

$\rm :\longmapsto\:h_m = \dfrac{ {u_0}^{2} {sin}^{2}\theta}{2g} - - - (1)$

and

The horizontal range of the projectile is given by

$\rm :\longmapsto\:R = \dfrac{ {u_0}^{2} {sin}2\theta}{g} - - - (2)$

Divide equation (1) by equation (2), we get

$\rm :\longmapsto\:\dfrac{h_m}{R} = \dfrac{\dfrac{ \cancel{ {u_0}^{2}} {sin}^{2} \theta}{2 \: \cancel{g}} }{\dfrac{ \cancel {u_0}^{2}sin2\theta}{ \cancel{g}}}$

$\rm :\longmapsto\:\dfrac{h_m}{R} = \dfrac{ {sin}^{2}\theta}{2 \: sin2\theta}$

$\rm :\longmapsto\:\dfrac{h_m}{R} = \dfrac{ {sin}^{2}\theta}{2 \times 2 \times sin\theta \times cos\theta}$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:sin2\theta = 2sin\theta \: cos\theta \bigg \}}$

$\rm :\longmapsto\:\dfrac{h_m}{R} = \dfrac{ {sin}\theta}{4cos\theta}$

$\rm :\longmapsto\:\dfrac{h_m}{R} = \dfrac{ {tan}\theta}{4}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:\dfrac{sin\theta}{cos\theta} = tan\theta \bigg \}}$

$\rm :\longmapsto\:\dfrac{4h_m}{R} = tan\theta$

$\rm :\longmapsto\: \: \theta \: = \: {tan}^{ - 1} \bigg(\dfrac{4h_m}{R} \bigg)$

Additional Information :-

Time of flight of projectile motion :-

$\rm :\longmapsto\:t \: = \: \dfrac{2u_0 \: sin\theta}{g}$

Answer 2

$\Huge\sf\mathbb\color{red}\underline{\colorbox{cyan}{Answer}}$

The maximum vertical height is given as,

$h_{m} = \frac{ u_{o} {sin}^{2} o_{o} }{2g}$

The horizontal range is given as,

$r = \frac{ u_{o} {sin}^{2} o_{o} }{g}$

From equation 1 and 2 ,we get

$\frac{ h_{m} }{r} = \frac{ {sin}^{2} o_{o}}{2 {sin}^{2} 2 o_{o} }$

$\frac{ h_{m} }{r} = \frac{sin \: o_{o}}{4 \cos }$

$tan \: o_{o} = \frac{4 h_{m} }{r}$

$o_{o} = {tan}^{ - 1} ( \frac{4 h_{m} }{r} )$

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