Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

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

Total time of ascent for a body is given 3 sec i.e. $t=\frac{usin\theta }{g}=3$

$\therefore usin\theta =10\times 3=30.............\left(i\right)$

Horizontal component of velocity remains always constant

$ucos\theta =vcos{30}^{\circ }..........\left(ii\right)$

For vertical upward motion between point O and A

$vsin{30}^{\circ }=usin\theta -g\times 2[Usingv=u-gtvsin{30}^{\circ }=30-20[Asusin\theta =30]$

Substituting this value in equation (ii) u coa $\theta =20cos{30}^{\circ }=10\sqrt{3}......\left(iii\right)$

From equation (i) and (iii) $u=20\sqrt{3}and\theta ={60}^{\circ }$