Question

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

Answer 1

Hint:In this question firstly, we will find the time, Maximum horizontal distance and the height of projectile and then accordingly we will put the values in the equation of maximum height to find out at what the range is to the maximum height.

Formula used:
Formula for range of a projectile
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where, $u$ is the speed and $g$ is the acceleration due to gravity.
Formula for height of a projectile
$H = \dfrac{{{u^2}{{\sin }^2}{\theta ^ \circ }}}{{2g}}$

Complete step by step answer:
Let us consider a body, which is projected with the speed $um{\text{ }}{s^{ - 1}}$ which is inclined at $\theta $ with the horizontal line.Then we can say that the vertical component of $u$ , ${u_y} = u\sin \theta $ and horizontal component of $u$ , ${u_x} = u\cos \theta $.Acceleration on vertical direction ${a_y} = - g$ and acceleration on horizontal direction ${a_x} = 0$. And we know that,
$x = {u_x}t \\
\Rightarrow x = u\cos \theta \times t \\ $
$ \Rightarrow t = \dfrac{x}{{u\cos \theta }}$ -----(1)
Similarly,
$y = {u_y}t + \dfrac{1}{2}{a_y}{t^2} \\
\Rightarrow y = u\sin \theta \times t + \dfrac{1}{2}( - g){t^2} \\ $
Now, substituting the value of $t$ in above equation,
$y = u\sin \theta \times \dfrac{x}{{u\cos \theta }} - \dfrac{1}{2}g \times {\left( {\dfrac{x}{{u\cos \theta }}} \right)^2} \\
\Rightarrow y = x\tan \theta - \dfrac{1}{2}g\dfrac{{{x^2}}}{{{u^2}{{\cos }^2}\theta }} \\ $
Which is as similar as the equation of a parabola $y = ax + b{x^2} + c$ .
As we know that, range equals to, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
And $R$ will be maximum, if
$\sin 2\theta = 1 \\
\Rightarrow 2\theta = {90^ \circ } \\
\Rightarrow \theta = {45^ \circ } \\ $
Then, ${R_{\max }} = \dfrac{{{u^2}}}{g}$
Therefore, the maximum height will be:
$H = \dfrac{{{u^2}{{\sin }^2}{{45}^ \circ }}}{{2g}}
\Rightarrow H = \dfrac{{{u^2}}}{g} \times \dfrac{1}{2}
\Rightarrow H = \dfrac{{{R_{\max }}}}{4} \\
\therefore {R_{\max }} = 4H \\ $
Hence, the range is 4 times the maximum height of the projectile.

Note:A projectile is a parabolic motion that is nothing but defined as the superposition of two fundamental forms of motion which are, constant acceleration in one direction and constant velocity in the other. Projectiles include thrown balls, rifle bullets, and falling bombs. Remember all the formulas of maximum height, time of flight and horizontal range of the projectile. It becomes very easy when we solve questions of this type.For easy calculation remember that the horizontal range is maximum when the angle of projection is ${45^ \circ }$ .