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 initial velocity are-

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

Answer 1

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


Time of ascent $=3$ sec$=$$\frac{T}{2}$
from time of flight
$\frac{2usin\theta }{g}=T$
$\frac{usin\theta }{g}=3sec$
$usin\theta =30\frac{m}{s}...\left(i\right)$
using
$V_y=u_y-gt$
$vsin{30}^{\circ }=usin\theta -10\times 2$
$vsin{30}^{\circ }=30-20=10$
$v=10\times 2=20\frac{m}{s}$
$ucos\theta =vcos{30}^{\circ }$
$\Rightarrow 20\times \frac{\sqrt{3}}{2}\Rightarrow 10\sqrt{3}{\text{ms}}^{-1}...\left(ii\right)$
both sides are squaring eq. (i) and (ii)
$u^2(si{n}^2\theta +co{s}^2\theta )=(30{)}^2+(10\sqrt{3}{)}^2$
$u^2=900+300$
$u=20\sqrt{3}{\text{ms}}^{-1}$
$tan\theta =\frac{30}{10\sqrt{3}}\Rightarrow \frac{3}{\sqrt{3}}={60}^{\circ }$