Answer to Question #198302 in Mechanics | Relativity for Sumit

Question

What is projectile? Show that path of the projectile is parabolic nature when it is projected from ground. 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

Projectile : A projectile is any object thrown by the exertion of a force. It can also be defined as an object launched into the space and allowed to move free under the influence of gravity and air resistance. Although any object in motion through space may be called projectiles, they are commonly found in warfare and sports.

Let a body is projected with speed $u\space m/s$  inclined $θ$  with horizontal line.

Then, vertical component of u, $u_y=u\sinθ$

Horizontal component of $u_x​=u\cosθ$

acceleration on horizontal, $a_x=0$

acceleration on vertical, $a_y=−g$

Now, use formula 

$x=u_x​t$

$x=u\cosθ.t$

$t=\dfrac{x}{u\cosθ}$ -------------------------(1)

Again, $y=u_y​t+\dfrac{1}{2}a_yt^2$

$y=u\sinθt−\dfrac{1}{2}gt^2$

put equation  (1) here,

$y=u\sinθ×\dfrac{x}{u\cosθ}−\dfrac{1}{2}g×\dfrac{x^2}{u^2\cos^2θ}$

$y=x\tanθ−\dfrac{1}{2}g\dfrac{x^2}{u^2\cos^2\theta}$

Which is similar to the equation of parabola,

$y=ax+bx^2+c$

We know that,

The range of the projectile $R=\dfrac{u^2\sin2θ}{g}$

​R will be the maximum, if $\sin2θ=1⇒2θ=90\degree$

$⇒θ=45\degree$

Then,

$R_{max}=\dfrac{u^2}g$

​

So the maximum height attained by the projectile is:

$H=\dfrac{u^2\sin^245\degree}{2g}=\dfrac{u^2}{g}​×\dfrac{1}{2}​=\dfrac{R_{max}}{4}$

$​​⇒R_{max}​=4H$