Question

Two seconds after projection, a projectile is traveling in a direction inclined at $30^\circ $ to the horizontal. After one more sec, it is traveling horizontally, the magnitude and direction of its velocity are

• A. $2\sqrt{2} \, ms^{- 1}, \, 60^\circ $
• B. $20\sqrt{3} \, ms^{- 1}, \, 60^\circ $
• C. $6\sqrt{40} \, ms^{- 1}, \, 30^\circ $
• D. $40\sqrt{6} \, ms^{- 1}, \, 30^\circ $

Answer 1

Answer: (B)

Let in 2s body reaches upto point A and after one more second upto point B.
Total time of ascent for a body is given as 3s

i.e., $\frac{u sin \theta }{g}=3$
$\therefore $ $usin \theta =10\times 3$ $=30$ ......(i)
Horizontal component of velocity remains always constant
$ucos \theta =vcos⁡30^\circ $ .......(ii)
For vertical upward motion between point O and A
$vsin 30^{o}=usin ⁡ \theta -g\times 2$ [Using v = u - gt]
$vsin 30^{o}=30-20$ $\left[A s \, u sin \theta = 30\right]$
$\therefore $ $v=20 \, ms^{- 1}$
Substituting this value $v$ in equation (ii), we get
$ucos \theta =20cos⁡30^\circ =10\sqrt{3}$ .....(iii)
From equations (i) and (ii)
$u=20\sqrt{3} \, $ $\text{m s}^{- 1}$ and $\theta =60^\circ $