Question

Trajectory of two particles projected from origin with speeds $v_1$ and $v_2$ at angles ${\theta }_1$ and ${\theta }_2$ with positive x-axis respectively are as shown in the figure. Given that $g=-10{\text{m/s}}^2\left(\hat{j}\right)$. Choose the correct option(s) related to diagram.

• A. $v_1-v_2=2v_1$
• B. ${\theta }_2-{\theta }_1=2{\theta }_1$
• C. $v_1=v_2$
• D. $3({\theta }_2-{\theta }_1)=+{\theta }_1$

Answer 1

Answer: (C), (D)

$3({\theta }_2-{\theta }_1)=+{\theta }_1$

Given that,
Initial speeds of two particles are $v_1$ and $v_2$, angles of projection are ${\theta }_1$ and ${\theta }_2$ respectively.
From the figure, we can deduce that
Maximum height of particle$-1$ $\left(H_1\right)=\frac{v_1^2{\sin }^2{\theta }_1}{2g}$
$\Rightarrow 10=\frac{v_1^2{\sin }^2{\theta }_1}{2g}.........\left(1\right)$
Range of particle$-1$ $\left(R_1\right)=\frac{v_1^2\sin 2{\theta }_1}{g}$
$\Rightarrow 40=\frac{v_1^2\sin 2{\theta }_1}{g}........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get

$\frac{1}{4}=\frac{\sin {\theta }_1}{4\cos {\theta }_1}\Rightarrow \frac{\tan {\theta }_1}{4}=\frac{1}{4}$

$\Rightarrow \tan {\theta }_1=1\Rightarrow {\theta }_1={45}^{\circ }$

Substituting this in equation $\left(2\right)$, we get, $v_1^2=400\Rightarrow v_1=20\text{m/s}$

Similarly, for second particle,

Maximum height of particle$-2$ $\left(H_2\right)=\frac{v_2^2{\sin }^2{\theta }_2}{2g}$
$\Rightarrow 15=\frac{v_2^2{\sin }^2{\theta }_2}{2g}.........\left(3\right)$
Range of particle$-2$ $\left(R_2\right)=\frac{v_2^2\sin 2{\theta }_2}{g}$
$\Rightarrow 20\sqrt{3}=\frac{v_2^2\sin 2{\theta }_2}{g}.......\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$ we get

$\frac{\sqrt{3}}{4}=\frac{\sin {\theta }_2}{4\cos {\theta }_2}\Rightarrow \tan {\theta }_2=\sqrt{3}\Rightarrow {\theta }_2={60}^{\circ }$

Substituting this in $\left(3\right)$ we get,

$15=\frac{v_2^2\times {\sin }^2{60}^{\circ }}{2\times 10}$

$\Rightarrow v_2^2=400\Rightarrow v_2=20\text{m/s}$

Therefore, we can conclude that, $v_1=v_2$ and $3({\theta }_2-{\theta }_1)=+{\theta }_1$.
Options (c) and (d) are correct.