Question

A particle is projected with speed $100\text{m/s}$ at angle $\theta ={60}^{\circ }$ with the horizontal at time $t=0$. At time ${}^{\prime }{t}^{\prime },$ the velocity vector of the particle becomes perpendicular to the direction of velocity of projection. Then,

• A. Its tangential acceleration at time ${}^{\prime }{t}^{\prime }$ is $5{\text{m/s}}^2$.
• B. Its radius of curvature at time ${}^{\prime }{t}^{\prime }$ is $\frac{2}{3\sqrt{3}}\text{km}$.
• C. Its tangential acceleration at time ${}^{\prime }{t}^{\prime }$ is $10{\text{m/s}}^2$.
• D. Its radius of curvature at time ${}^{\prime }{t}^{\prime }$ is $\frac{2}{2\sqrt{3}}\text{km}$.

Answer 1

Answer: (A), (B)

The correct options are
A Its tangential acceleration at time ${}^{\prime }{t}^{\prime }$ is $5{\text{m/s}}^2$.
B Its radius of curvature at time ${}^{\prime }{t}^{\prime }$ is $\frac{2}{3\sqrt{3}}\text{km}$.


As initial angle of projection is ${60}^{\circ }$, angle made by velocity vector with the horizontal when perpendicular to initial velocity is ${30}^{\circ }$

Component of $g$ along path of projectile is
$a_T=g\cos {60}^{\circ }=10\times \frac{1}{2}=5{\text{m/s}}^2$
$a_n=\frac{v^2}{R}=g\sin {60}^{\circ }$
As, horizontal velocity remains same
$100\cos {60}^{\circ }=v\cos {30}^{\circ }$
$v=\frac{50}{\sqrt{3}}\times 2=\frac{100}{\sqrt{3}}\text{m/s}$
Radius of curvature
$R=\frac{v^2}{a_n}=\frac{v^2}{g\sin 60}=\frac{100\times 100\times 2}{3\times 10\times \sqrt{3}}$
$=\frac{2000}{3\sqrt{3}}\text{m}$
$=\frac{2}{3\sqrt{3}}\text{km}$