Question

A particle is projected at an angle $\theta$ from ground with speed $u(g=10m/s^2)$, then which of the following is true?

• A. If $u=10m/s$ and $\theta ={30}^o$, then time of flight will be 1 sec
• B. If $u=10\sqrt{3}m/s$ and $\theta ={60}^o$, then time of flight will be 3 sec
• C. If $u=10\sqrt{3}m/s$ and $\theta ={60}^o$, then after 2 sec velocity becomes perpendicular to initial velocity
• D. If $u=10m/s$ and $\theta ={30}^o$, then velocity never becomes perpendicular to intial velocity during its flight

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A If $u=10\sqrt{3}m/s$ and $\theta ={60}^o$, then time of flight will be 3 sec
B If $u=10m/s$ and $\theta ={30}^o$, then time of flight will be 1 sec
C If $u=10\sqrt{3}m/s$ and $\theta ={60}^o$, then after 2 sec velocity becomes perpendicular to initial velocity
D If $u=10m/s$ and $\theta ={30}^o$, then velocity never becomes perpendicular to intial velocity during its flight

Using the formula,

$T=\frac{2uSin\theta }{g}$

For OPTION A, we have,

$u=10m/s$

$\theta =30{}^{\circ }$

$T=\frac{2\times 10\times Sin30}{10}$

$T=1sec$

For OPTION B, we have,

$u=10\sqrt{3}m/s$

$\theta =60{}^{\circ }$

$T=\frac{2\times 10\sqrt{3}\times Sin60}{10}$

$T=3sec$

Time taken by velocity to become perpendicular to the initial velocity is given by:

$t=\frac{u}{gSin\theta }$

For OPTION C, we have,

$u=10\sqrt{3}m/s$

$\theta =60{}^{\circ }$

$t=\frac{10\sqrt{3}}{10\times Sin60}$

$t=2sec$

For OPTION D, we have,

$u=10m/s$

$\theta =30{}^{\circ }$

$t=\frac{10}{10\times Sin30}$

$t=2sec$

Since, $T=1sec\le t=2sec$

$\therefore$ The velocity never becomes perpendicular to the initial velocity during its flight as time of flight is less than time when velocity becomes perpendicular to the initial velocity.

Hence, OPTIONS A, B, C, D are correct.