Question

The velocity of a projectile at the initial point $A$ is $(2\hat{i}+3\hat{j})\text{m/s}$. Its velocity (in m/s) at point $B$ is

• A. $-2\hat{i}+3\hat{j}$
• B. $2\hat{i}+3\hat{j}$
• C. $-2\hat{i}-3\hat{j}$
• D. $2\hat{i}-3\hat{j}$

Answer 1

The correct option is D $2\hat{i}-3\hat{j}$

Given, initial velocity at point $A$, $\overrightarrow{u}=(2\hat{i}+3\hat{j})\text{m/s}$.
$\overrightarrow{u}=(u_x\hat{i}+u_y\hat{j})\text{m/s}$
where $u_x$ and $u_y$ is the velocity of the particle along $x$ and $y-$ axis.

Along $y-$ axis-
$u_y=3\text{m/s}$
There is gravitational force acting downward. So, first speed will decrease and become zero at maximum height and then it will increase in the downward direction.
At $A$ and $B$, heights are same, so by law of conservation of mechanical energy, magnitude of velocities at $A$ and $B$ will be same but directions are opposite
$\therefore v_y=-3\text{m/s}$

Along $x-$ axis-
$u_x=2\text{m/s}$
There is no force acting in $x-$ direction. Thus, there will be no change in velocity.
$\therefore v_x=2\text{m/s}$
Hence, final velocity of the projectile at $B$ is
$(2\hat{i}-3\hat{j})\text{m/s}$.

$\begin{array}{c}\text{Why this question ?}\\ \text{It insists you to apply your concept of force}\\ \text{and law of conservation of mechanical energy}\\ \text{without using any calculations.}\end{array}$