Question

A projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80 \,m \,s ^{-1}$. Its time of flight is T Its velocity at the moment $t=\frac{T}{4},$ has a magnitude of nearly (Take $\left.g=10 \,m \,s ^{-2}\right)$

• A. $200 \,m \,s ^{-1}$
• B. $160 \,m \,s ^{-1}$
• C. $144 \,m \,s ^{-1}$
• D. $140\, m \,s ^{-1}$

Answer 1

Answer: (C)

Vertical component of initial velocity,
$u_{y}=u \,\sin \,30^{\circ}$
or $u=\frac{u_{y}}{\sin 30^{\circ}}=\frac{80}{(1 / 2)}=160 \,m\, s ^{-1}$
Horizontal component of initial velocity,
$u_{x}=u \cos 30^{\circ}=160 \times \frac{\sqrt{3}}{2}=80 \sqrt{3} m s ^{-1}$
Time of flight, $T=\frac{2 u \sin \theta}{g}$
$=\frac{2 \times 160 \times \sin 30^{\circ}}{10}=16 s$
$\therefore t=\frac{T}{4}=\frac{16}{4}=4 s$
Let $v$ be the velocity of the projectile at $t=\frac{T}{4}$
Its horizontal and vertical components are given by
$v_{x}=u_{x}=80 \sqrt{3} m s ^{-1} $
$v_{y}=u_{y}-g t=80-10 \times 4=40 m s ^{-1}$
Its magnitude is given by
$v =\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(80 \sqrt{3})^{2}+(40)^{2}}$
$=40 \sqrt{12+1}=40 \sqrt{13} \approx 144 \,m \,s ^{-1}$