Question

A body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40\, ms^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly, (Acceleration due to gravity = $10 \,ms^{-2}$)

• A. $3.54 \; ms^{-1}$
• B. $35.40 \; ms^{-1}$
• C. $30.54 \; ms^{-1}$
• D. $34.5 \; ms^{-1}$

Answer 1

Answer: (C)

According to the question,

Verticle component of velocity,
$u_{y}=\frac{\sqrt{3}}{2} u=40$
or $u=\frac{80}{\sqrt{3}} ms ^{-1}$
Time of flight, $T=\frac{2 u \sin \theta}{g}$
$=\frac{2\left(\frac{80}{\sqrt{3}}\right) \sin 60^{\circ}}{10} $
$ \Rightarrow T =2 \times \frac{80}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \times \frac{1}{10}$ $\Rightarrow T=8 sec$
at time, $t=\frac{T}{4}=\frac{8}{4}=2 sec$
$v_{x}=\frac{\frac{80}{\sqrt{3}}}{2}=\frac{40}{\sqrt{3}} ms ^{-1}$
at $t=\frac{T}{4}=2 sec$,
From first equation of the motion,
$v_{y}=u_{y}-g t=\frac{\sqrt{3}}{2}\left(\frac{80}{\sqrt{3}}\right)-10 \times 2$
${\left[\because u_{y}=u \sin 60^{\circ}\right]}$
$=40-10 \times 2=20 ms ^{-1}$
Hence, at $t=\frac{T}{4}$, magnitude of velocity of projectile,
$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$
$=\sqrt{\left(\frac{40}{\sqrt{3}}\right)^{2}+(20)^{2}} \approx 30.54 ms ^{-1}$