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

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., $\frac{usin\theta }{g}=3$
$\therefore$ $usin\theta =10\times 3$ $=30$ ......(i)
Horizontal component of velocity remains always constant
$ucos\theta =vcos30^{\circ }$ .......(ii)
For vertical upward motion between point O and A
$vsin30^o=usin\theta -g\times 2$ [Using v = u - gt]
$vsin30^o=30-20$ $\left[Asusin\theta =30\right]$
$\therefore$ $v=20m{s}^{-1}$
Substituting this value $v$ in equation (ii), we get
$ucos\theta =20cos30^{\circ }=10\sqrt{3}$ .....(iii)
From equations (i) and (ii)
$u=20\sqrt{3}$ ${\text{m s}}^{-1}$ and $\theta =60^{\circ }$