Question

prove that the maximum horizontal range is 4 times the maximum height attained by the projectile time when fired at the inclination so as to have maximum horizontal range

Answer 1

let's take the equation of trajectory of projectile
$y = \tan( \alpha ) x - (\frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} } ) {x}^{2}$
then taking this equation with respect to x-axis we get
substitute x = r & y = 0 we get

$0 = \tan( \alpha ) r - ( \frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} } ) {r}^{2}$
then shifting negative term to L.H.S we get

$( \frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} } ) {r}^{2} = \tan( \alpha ) r$
then after simplification we get

$(\frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} } )r = \tan( \alpha )$
$\frac{1}{2} ( \frac{gr}{ {u}^{2} { \cos( \alpha ) }^{2} } ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$
then keeping 'r' on L.H.S and shifting all the terms on R.H.S we get

$r = \frac{ {u}^{2}2 \sin( \alpha ) { \cos( \alpha ) }^{2} }{g \cos( \alpha ) }$
$r = \frac{ {u}^{2}2 \sin( \alpha ) \cos( \alpha ) }{g}$
then applying sin double angle formula we get
sin2(angle) = 2sincos

$r = \frac{ {u}^{2} \sin(2 \alpha ) }{g}$
Consider an object is projected i.e (thrown in the air & moving freely under action of gravity) with some initial velocity 'u' & making angle of projection with respect to ground we get
$\alpha = {45}^{o}$
$r = \frac{ {u}^{2} \sin2 \: ({45}^{o} ) }{g}$
$r = \frac{ {u}^{2} \sin(90) }{g} \\ or \\ r = \frac{ {u}^{2} 2 \sin(45) \cos(45) }{g}$
as sin90° = 1 &
sin45° = 1/√2 as well as cos45° = 1/√2
therefore,

$r = \frac{ {u}^{2}(1) }{g} \\ or \\ r = \frac{ {u}^{2} \frac{1}{ \sqrt{2} }( \frac{1}{ \sqrt{2} } ) }{g}$
$r {max}= \frac{ {u}^{2} }{g}$
it means a projectile has maximum range when it is projected at an angle of 45°

let's consider an object is projected with initial velocity (u) making some angle with respect to ground level i.e angle of projection so it's height during the motion would be (h) and this height of the projectile during the motion is given by

${v}^{2} = {u}^{2} + 2as$
then taking this kinematical equation with respect to y-axis we get
substitute v = 0, u = sin, a = g & s = h

$0 =u^2 { \sin( \alpha ) }^{2} - 2gh$
then shifting negative term on L.H.S we get

$2gh =u^2 { \sin( \alpha ) }^{2}$
then after simplification we get formula for maximum height of projectile during the motion

$h = \frac{ { u^2\sin( \alpha ) }^{2} }{2g}$
then substitute angle 45° we get


h(max) = u^2/4g There after h(max) = u^2/4g Therefore, h(max) = u^2/g × 1/4 i.e. h(max) = Rmax/4 Rmax = 4Hmax

Answer 2

Answer is in the image