Question

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

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

Answer 1

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

horizontal component at $t=2s$ is $vcos{30}^o$ and this will remain constant through out the motion

vertical component is $usin30$

$v=u+at$

$vsin30=usin\theta -2g$

$v=2usin\theta -4g=2g$

time to reach the top is $3s$ therefore

$v=u+at$

$0=usin\theta -3g$

$usin\theta =3g$

$ucos\theta =vcos{30}^o=2g\times \frac{\sqrt{3}}{2}=\sqrt{3}g$

$u=\sqrt{u_x^2+u_y^2}=20\sqrt{3}m/s$

$\theta =ta{n}^{-1}\frac{u_y}{u_x}=ta{n}^{-1}\frac{3g}{\sqrt{3}g}=ta{n}^{-1}\sqrt{3}={60}^o$