Question

A body is projected at ${30}^{\circ }$ with velocity $v_1$ from point $A$ on the ground. At the same time another body is projected vertically upwards from $B$ with velocity $v_2$ from the ground. The point $B$ lies vertically below the highest point of the trajectory of ball $A$. For both the bodies to collide, what should be the ratio of $v_1/v_2$ ?

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

Answer 1

The correct option is D $2$

For balls to collide, vertical height covered by ball $A$ must be equal to the vertical height covered by ball $B$ in same time.
Let $T$ be the time of flight of a projectile for ball $A$.
The point where ball $B$ collides with ball $A$ is also the highest point of trajectory of ball $B$.
Let $H$ be the maximum height of the projectile of ball $A$.
Time taken by ball $A$ to reach maximum height of the projectile = $\frac{T}{2}$
For Ball $A$:
$H=v_1\sin {30}^{\circ }\left(\frac{T}{2}\right)-\frac{1}{2}g{\left(\frac{T}{2}\right)}^2.....\text{(i)}$ (Using second equation of motion)
For Ball $B$:
$H=v_2\left(\frac{T}{2}\right)-\frac{1}{2}g{\left(\frac{T}{2}\right)}^2.....\text{(ii)}$
On comparing $\text{(i)}$ and $\text{(ii)}$, we get
$\frac{v_1}{2}=v_2$
$\Rightarrow \frac{v_1}{v_2}=2$