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

Answer 1

Answer: (B)

$20\sqrt{3}m ⥂ / ⥂ {sec,}{}60^{o}$

Answer 2

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

$\therefore u\sin\theta = 10 \times 3 = 30$ …..(i)

Horizontal component of velocity remains always constant

$u\cos\theta = v\cos 30{^\circ}$ …..(ii)

For vertical upward motion between point O and A

$v\sin 30^{o} = u\sin\theta - g \times 2$ $\left\lbrack \text{Using }v = u - gt \right\rbrack$

$v\sin 30^{o} = 30 - 20$

$\left\lbrack \text{As}u\sin\ \theta = \text{30} \right\rbrack$

$\therefore v = 20m/s.$

Substituting this value in equation (ii)

$u\cos\theta = 20\cos 30^{o} = 10\sqrt{3}$ …..(iii)

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