Question

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

• A. $3.54m{s}^{-1}$
• B. $35.40m{s}^{-1}$
• C. $30.54m{s}^{-1}$
• D. $34.5m{s}^{-1}$

Answer 1

Answer: (C)

According to the question,

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