Question

If $T$ is the total time of flight, $H$ is the maximum height and $R$ is the horizontal range of a projectile, then $x$ and $y$ coordinates at any time $t$ are related as (Take point of projection as origin)

• A. $y=4H\left(\frac{R}{x}\right)[1-\frac{R}{x}]$
  
• B. $y=4H\left(\frac{x}{R}\right)[1-\frac{x}{R}]$
  
• C. $y=4H\left(\frac{T}{t}\right)[1-\frac{T}{t}]$
  
• D. $y=4H\left(\frac{t}{T}\right)[1-\frac{t}{T}]$
  

Answer 1

Answer: (B), (D)

$y=4H\left(\frac{t}{T}\right)[1-\frac{t}{T}]$


As we know that for a projectile motion,

Total time of flight, $T=\frac{2u\sin \theta }{g}$

Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$

Horizontal range, $R=\frac{u^2\sin 2\theta }{g}$

Now,
$\frac{H}{R}=\frac{u^2{\sin }^2\theta }{2g}\times \frac{g}{2u^2\sin \theta \cos \theta }$

$\Rightarrow \frac{H}{R}=\frac{\tan \theta }{4}$

$\Rightarrow \tan \theta =\frac{4H}{R}$

Using maximum height formula we can write that,

$u^2=\frac{2gH}{{\sin }^2\theta }$

Equation of trajectory of projectile gives,
$$y=\left(\tan \theta \right)x-\frac{g}{2u^2{\cos }^2\theta }{x}^2$$
Substituting the value of $u^2$, we get
$\Rightarrow y=\left(\tan \theta \right)x-\frac{g{x}^2}{2\left(\frac{2gH}{{\sin }^2\theta }\right){\cos }^2\theta }$

$\Rightarrow y=\left(\tan \theta \right)x-\frac{{\tan }^2\theta }{4H}{x}^2$

$\Rightarrow y=\frac{4H}{R}x-\frac{1}{4H}{\left(\frac{4H}{R}\right)}^2{x}^2\because [\tan \theta =\frac{4H}{R}]$

$\Rightarrow y=\frac{4H}{R}[x-\frac{x^2}{R}]$

$\therefore y=4H\left(\frac{x}{R}\right)[1-\frac{x}{R}]$

So, Option (b) is the correct answer.

The vertical displacement $y$ is,
$y=u\sin \theta t-\frac{1}{2}g{t}^2.....\left(1\right)$

Now, $\frac{H}{T}=\frac{\left(\frac{u^2{\sin }^2\theta }{2g}\right)}{\left(\frac{2u\sin \theta }{g}\right)}$

$\Rightarrow \frac{H}{T}=\frac{u\sin \theta }{4}$

$\therefore \frac{4H}{T}=u\sin \theta$

Also, $\frac{H}{T^2}=\frac{u^2{\sin }^2\theta }{2g}\times \frac{g^2}{4u^2{\sin }^2\theta }$

$\Rightarrow \frac{H}{T^2}=\frac{g}{8}$

$\therefore g=\frac{8H}{T^2}$

Substituting $u\sin \theta$ and $g$ in equation $\left(1\right)$, we get

$y=\frac{4H}{T}t-\frac{1}{2}\times \frac{8H}{T^2}\times t^2$

$\therefore y=4H\left(\frac{t}{T}\right)[1-\frac{t}{T}]$

Hence options (b) and (d) are the correct alternatives.