Question

A particle is projected in the X-Y plane. $2s$ after projection, the velocity of the particle makes an angle $45^{\circ }$ with the X-axis. $4s$ after projection, it moves horizontally. Find the velocity of projection (use $g=10m{s}^{-2}$ )

• A. $20\sqrt{5}{\text{m s}}^{-1}$
• B. $10\sqrt{5}{\text{m s}}^{-1}$
• C. $5\sqrt{5}{\text{m s}}^{-1}$
• D. $2\sqrt{5}{\text{m s}}^{-1}$

Answer 1

Answer: (A)

$v_y=u_y+a_{y}t$
at the highest point of motion, $v_y=0$
$0=u_y-gt$
$u_y=4g=40m{s}^{-1}$


at $t=2s$
$v_y=u_y+a_{y}t$
$v_y=4g-gt=20m{s}^{-1}$
given that $v_y=u_x=20m{s}^{-1}$
the velocity of projection $u=\sqrt{{\text{u}}_{\text{x}}^2+{\text{u}}_{\text{y}}^2}$
$\Rightarrow \text{u}=\sqrt{{\left(40\right)}^2+{\left(20\right)}^2}$
$\text{u}=\sqrt{2000}\Rightarrow 20\sqrt{5}{\text{m s}}^{-1}$