Question

A particle is projected with speed $V_A$ from a point making an angle ${60}^0$ with horizontal. At the same instant, a second particle B is thrown vertically upward from point directly below the maximum height point of parabolic path of A with velocity $V_B$. If the two particles collide then the ratio of $V_A/V_B$ should be:-

• A. $1$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}$

$\text{Step 1: Condition of collision}$

Clearly from figure, the particles can collide at the maximum height of the projectile.

Let the particles collide in time $t$.

Since the projectile motion is symmetric, So time to reach maximum height $t=\frac{T}{2}$, Where $T$ is the time of flight of the projectile.

Till collision, distance travelled by B equals the Maximum height of the projectile, i.e. $h$

$\text{Step 2 : Maximum height and Time Calculation}$

Taking upward as positive

Maximum height travel by the particle $A$ is

$\begin{array}{c}h=\frac{u^{2}si{n}^2\theta }{g}=\frac{V_A^2{\sin }^2{60}^o}{2g}\\ =\frac{3V_A^2}{8g}\end{array}$ $....\left(1\right)$

$\text{Step 3 : Distance traveled by the second particle in time t}$

For Particle A:

Time of flight $T=\frac{2usin\theta }{g}=\frac{2V_A\sin {60}^o}{g}$

Now, $t=T/2=\frac{V_A\sin {60}^o}{g}=\frac{V_A\sqrt{3}}{2g}$

For Particle B:

Since acceleration is constant, therefore applying equation of motion in y direction, (Taking upward as positive)

$\begin{array}{c}S=V_{B}t-\frac{1}{2}g{t}^2\\ =V_B\left(\frac{V_A\sqrt{3}}{2g}\right)-\frac{1}{2}g\times \frac{V_A^2\times 3}{4g^2}\end{array}$ $....\left(2\right)$

$\text{Step 3: Calculations}$

Since distance traveled by B equal the maximum height of projectile, So using Equations $\left(1\right)$ and $\left(2\right)$

$\therefore S=h$

$\begin{array}{c}\Rightarrow V_B\left(\frac{V_A\sqrt{3}}{2g}\right)-\frac{1}{2}g\times \frac{V_A^2\times 3}{4g^2}=\frac{3V_A^2}{8g}\\ \Rightarrow \frac{V_A}{V_B}=\frac{2}{\sqrt{3}}\end{array}$

Hence, Option $B$ is correct.