Question

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

• 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,
$u=140\, ms ^{-1}$
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 $v_{y} =u_{y}-g t$
$140 \,\cos \,60^{\circ} =140\, \sin\, 60^{\circ}-10 t$
$ \left[\because v_{y}=v_{x}\right]$
$ 140 \times \frac{1}{2} =140 \times \frac{\sqrt{3}}{2}-10 t$
$ 70 =70 \sqrt{3} \, 10 t $
$ 10 t =70 \sqrt{3}-70$
$ t =\frac{70(\sqrt{3}-1)}{10} $
$=7 \times(0.7320)=5.124\, s $