Question

A particle is fired at an angle of ${30}^{\circ }$ to the horizontal such that the vertical component of its initial velocity is $80m/s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly
(Take $g=10m/s^2$)

• A. $200m/s$
• B. $300m/s$
• C. $144m/s$
• D. $100m/s$

Answer 1

The correct option is C $144m/s$

$u_y=80m/s=u\sin {30}^{\circ }$
$\Rightarrow u=160m{s}^{-1}\therefore u_x=u\cos {30}^{\circ }=160\times \frac{\sqrt{3}}{2}=80\sqrt{3}m{s}^{-1}$
Time of flight $T=\frac{2u\sin \theta }{g}=\frac{2\times 80}{10}=16s$
At $t=\frac{T}{4}=4s,$
$v_x=u_x=80\sqrt{3}m{s}^{-1}$ and $v_y=u_y-gt=80-10\times 4=40m{s}^{-1}$
$v=\sqrt{v_x^2+v_y^2}=\sqrt{(80\sqrt{3}{)}^2+(40{)}^2}=40\sqrt{12+1}\therefore v=40\sqrt{13}=144m{s}^{-1}$