Question

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

• 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, $v_{y}=0$
$0=u_{y}-gt$
$u_{y}=4g=40ms^{- 1}$


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