Question

If time taken by the projectile to reach $\text{Q}$ is $T$, then what is the magnitude of $\text{PQ}$?

• A. $Tv\sin \theta$
• B. $Tv\cos \theta$
• C. $Tv\sec \theta$
• D. $Tv\tan \theta$

Answer 1

The correct option is D $Tv\tan \theta$

Considering $\text{X}$ and $\text{Y}$ axes as shown in the figure.


According to given problem,

velocity along the $\text{X-axis}$, $U_x=0$

velocity along the $\text{X-axis}$, $U_y=v$

For $\text{P}\to \text{Q}$,

Range on $\text{X-axis}=\text{PQ}$

Net displacement along $\text{Y-axis}$

$Y_{net}=0$

Using equation of motion,$$Y_{net}=U_{y}t+\frac{1}{2}{a}_y{t}^2$$Substituting $a_y=g\cos \theta$ from the figure and the other given data, we get

$0=vT-\frac{1}{2}g\cos \theta .T^2$

$\Rightarrow T=\frac{2v}{gcos\theta }$

$\Rightarrow g=\frac{2v}{Tcos\theta }$

In the same time $T$ displacement along $\text{X-axis}$ is $\text{PQ}$.

Applying equation of motion along $\text{X-axis}$, we have

$PQ=U_{x}t+\frac{1}{2}{a}_x{t}^2$

Substituting $a_x=g\sin \theta$ and the other parameters given in the problem,

$PQ=0+\frac{1}{2}g\sin \theta T^2$

$\Rightarrow PQ=\frac{1}{2}g\sin \theta T^2$

Substituting the value of $g$,

$PQ=\frac{1}{2}\frac{2v}{T\cos \theta }\sin \theta T^2$

$\therefore PQ=Tv\tan \theta$

Hence, option (d) is the correct answer.