Question

A projectile is launched with an initial velocity $\overrightarrow{V_0}=\left(2m/s\right)\hat{i}+\left(3m/s\right)\hat{j}$. At the top of the trajectory, the speed of the particle is ($x$ horizontal direction, $y$-vertical dirction)-

• A. $\sqrt{2^2+3^2}m/s$
• B. $2m/s$
• C. $3m/s$
• D. $5m/s$

Answer 1

The correct option is B $2m/s$

Given,

$v_0=\left(2m/s\right)\hat{i}+\left(3m/s\right)\hat{j}$

At the top of the trajectory, the vertical component of the speed gets vanished.

So, $v_y=0m/s$

And the Horizontal component of the speed,

$v_x=2m/s$

At the top of the trajectory, speed of the particle is

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\overrightarrow{v}=2\hat{i}+0\hat{j}$

$v=|\overrightarrow{v}|=\sqrt{(2{)}^2+(0{)}^2}$

$v=2m/s$

The correct option is B.