Question

A projectile is thrown in the upward direction making an angle of $60^{\circ }$ with the horizontal with a velocity of $140m{s}^{-1}$. Then, the time after which its velocity makes $45^{\circ }$ with the horizontal is (Acceleration due to gravity = $10m{s}^{-2}$ )

• A. 0.5124 s
• B. 51.24 s
• C. 5.124 s
• D. 512.4 s

Answer 1

Answer: (C)

Given, angle of projection, $\theta =60^{\circ }$ and velocity,
$u=140m{s}^{-1}$
This velocity have divided into two components,
horizontal component, $u_x=u\cos 60^{\circ }=140\cos 60^{\circ }$
and vertical component, $u_y=u\sin 60^{\circ }=140\sin 60^{\circ }$
Let after time $t$, the inclination of particle with horizontal be $45^{\circ }$ and at time $t$ velocity along
$x=v_x$ and along $y=v_y$.
Now, $\tan 45^{\circ }=\frac{v_y}{v_x}$
$=v_x=v_y$
Since, the horizontal component of velocity remains constant, i.e.,
$v_x=u_x=140\cos 60^{\circ }$
and $v_y=u_y-gt$
$140\cos 60^{\circ }=140\sin 60^{\circ }-10t$
$\left[\because v_y=v_x\right]$
$140\times \frac{1}{2}=140\times \frac{\sqrt{3}}{2}-10t$
$70=70\sqrt{3}10t$
$10t=70\sqrt{3}-70$
$t=\frac{70(\sqrt{3}-1)}{10}$
$=7\times (0.7320)=5.124s$