Question

Show that the maximum range of a projectile in any direction is described in the same time in which it would fall freely under gravity through this distance starting from rest.

Answer 1

To Prove:

Maximum range of a projectile is described in the same time in which it takes to fall freely under gravity through the same distance starting from rest.

Proof:

For Projectile:

Let range be R , initial velocity of projectile be u , gravity be g and $\theta$ be angle of Projection.

$\therefore \: R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

For maximum value of R , the value of sine function should be maximum

$\therefore \: \sin(2 \theta) = 1$

$= > 2 \theta = 90 \degree$

$= > \theta = 45 \degree$

So maximum range is obtained at 45° angle of Projection.

Time taken for maximum range be T

$\therefore \: T = \dfrac{2u \sin( \theta) }{g}$

$= > \: T = \dfrac{2u \sin( 45 \degree) }{g}$

$= > \: T = \dfrac{2u \times ( \frac{1}{ \sqrt{2} }) }{g}$

$= > T = \dfrac{u \sqrt{2} }{g}$

For Vertical Free Fall:

Height = Max range = u²/g

$\therefore \: H = \dfrac{1}{2} g {t}^{2}$

$= > \dfrac{ {u}^{2} }{g} = \dfrac{1}{2} g {t}^{2}$

$= > {t}^{2} = \dfrac{2 {u}^{2} }{ {g}^{2} }$

$= > t = \dfrac{u \sqrt{2} }{g}$

So , in both cases , t and T are same

[Hence Proved]

Answer 2

Question:-

Show that the maximum range of a projectile in any direction is described in the same time in which it would fall freely under gravity through this distance starting from rest.

Answer:-

$\bold{Finding \:\theta\: for\: which\: range\: is\: maximum:-}$

$\sf{R\:=\:\dfrac{{u}^{2}sin(2\theta)}{g}}$

R is maximum when $\bf{sin(2\theta)}$ is maximum

$\bf{sin(90)}$ = 1 is maximum value possible

Then $\bf{2\theta }$ = 90°

So $\bf{\theta}$ = 45°

Range:-

$\sf{R\:=\:\dfrac{{u}^{2}sin(2\theta)}{g}}$

$\sf{\implies\:R\:=\:\dfrac{{u}^{2}1}{g}}$

$\sf{\implies\:R\:=\:\dfrac{{u}^{2}}{g}}$

Time of flight:-

$\bf{Time\:of\:flight\:for\:\theta\:=\:45°:-}$

$\sf{T\:=\:\dfrac{2usin(\theta)}{g}}$

$\sf{\implies\:T=\dfrac{2u*\dfrac{1}{\sqrt{2}}}{g}}$

$\sf{\implies\:=\:T=\dfrac{\sqrt{2}u}{g}}$

$\rule{400}{4}$

Vertical free fall:-

Given: Height = max range = u²/g

Height = H

Apply s = ut + 1/2(a)(t²)

=> -H = 0 - 1/2(g)(t²)

=> H = 1/2(g)t²

Given: H = u²/g

=> $\sf{\dfrac{{u}^{2}}{g}\:=\:\dfrac{1}{2}(g){t}^{2}}$

=> $\sf{\dfrac{2{u}^{2}}{{g}^{2}}\:=\:{t}^{2}}$

=> $\sf{t\:=\:\sqrt{\dfrac{2{u}^{2}}{{g}^{2}}}}$

=> $\sf{t\:=\:\dfrac{\sqrt{2}u}{g}}$

T = t

Hence, proved