Question

A particle is projected from a point P with a velocity $v$ at an angle $\theta$ with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

• A. velocity of particle at Q is $v\sin \theta$
• B. velocity of particle at Q is $v\cot \theta$
• C. time of flight from P to Q is $\frac{v}{g}$ $\text{cosec}\theta$
• D. time of flight from P to Q is $\frac{v}{g}\sec \theta$

Answer 1

Answer: (B), (C)

velocity of particle at Q is $v\cot \theta$
time of flight from P to Q is $\frac{v}{g}$ $\text{cosec}\theta$

Let net velocity at Q is $v_q$, it makes an $(\theta -90)$ with horizontal as it is vertical to $v_q$ and $v_p$ makes $\theta$ with horizontal .

Now as there is no acceleration along x and air resistance is ignored , horizontal component of velocity remains unchanged .

i.e $v_p\cos \theta$ = $v_q\cos (\theta -90)$

$v_q$ = $v_p\cot \theta$ = $v\cot \theta$ ..... (i)

initial vertical velocity +$v\sin \theta$

Final vertical component of velocity

= $v_q\sin (\theta -90)$ =$-v\cot \theta \times \cos \theta$ = $-v({\cos \theta )}^2/\sin \theta$

Using laws of motion :

$-\frac{v({\cos \theta )}^2}{\sin \theta }=v\sin \theta -gt$

$t$=$vcosec\theta /g$ .... (ii) (Time of flight from P to Q )