Question

Two seconds after projection, a projectile is travelling in a direction inclined at ${30}^o$ to the horizontal. After one more sec, it is travelling horizontally, the magnitude and direction of its velocity are :

• A. $2\sqrt{20}m/s,{60}^o$
• B. $20\sqrt{3}m/s,{60}^o$
• C. $6\sqrt{40}m/s,{30}^o$
• D. $40\sqrt{6}m/s,{30}^o$

Answer 1

The correct option is B $20\sqrt{3}m/s,{60}^o$

Let in $2s$ body reaches upto point $A$ and after one more sec upto point $B$

Total time of ascent for a body is given $3s$

i.e., $t=\frac{u\sin \theta }{g}=3$

$\therefore u\sin \theta =10\times 3$

$u\sin \theta =\mathrm{30...}\left(i\right)$

Horizontal component of velocity remains always constant

$u\cos \theta =v\cos {30}^o...\left(ii\right)$

for vertical upward motion between point $O$ and $A$

$v\sin {30}^o=u\sin \theta -g\times 2=30-20$ (Using $v=u-gt$)

$\therefore v=20m/s$

Substituting this value $v$, we get Eq. (ii)

$u\cos \theta =20\cos {30}^o...\left(iii\right)$

from Eqs (i) and (iii)

$u=20\sqrt{3}m/s,{;\theta =60}^o$