Question

A particle is projected from ground with velocity $40\sqrt{2}m/sat{45}^{\circ }.$ Find the velocity after $2$ sec.

• A. $40\sqrt{5}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with horizontal
• B. $20\sqrt{5}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with horizontal
• C. $40\sqrt{3}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with horizontal
• D. $20\sqrt{3}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with horizontal

Answer 1

The correct option is B $20\sqrt{5}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with horizontal

$\text{Step 1: Initial velocities in x and y direction}$$\text{[Ref. Fig. 1]}$

Resolve $u$ along horizontal and vertical direction.

Initial velocity in x and y direction :

$u_x=u\cos \theta$ $=40\sqrt{2}\cos {45}^{\circ }m/s$ $=40m/s$

$u_y=u\sin \theta$$=40\sqrt{2}\sin {45}^{\circ }m/s$$=40m/s$

$\text{Step 2: Final velocities in x and y direction}$ $\text{[Ref. Fig 2]}$

In x direction: At $t=2s$, Velocity in $x$ remains same as $a_x=0$

In y direction: At $t=2s$, $a_y=-10m/s^2$ , $u_y=40\sqrt{2}\sin {45}^o=40m/s$

Since acceleration is constant, therefore applying equation of motion

$V_y=u_y+a_{y}t$

$=40m/s-10m/s^2\times 2s$$=20m/s$

$V_x=u_x=40m/s$

$\text{Step 3: Resultant Velocity}$ $\text{[Ref. Fig 2]}$

Resultant Velocity of the particle after 2s:

$V=\sqrt{V_x^2+V_y^2}$

$=\sqrt{{40}^2+{20}^2}m/s$$=20\sqrt{5}m/s$

From figure: $\tan \theta =\frac{V_y}{V_x}=\frac{20}{40}=\frac{1}{2}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{2}\right)$

Hence velocity after $2s$ is $20\sqrt{5}m/s$ at angle ${\tan }^{-1}\left(\frac{1}{2}\right)$. Option $B$ is correct.