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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

velocity of particle at Q is $v \cot θ$

velocity of particle at Q is $v \sin θ$

time of flight from P to Q is $(v / g) s e c θ$

time of flight from P to Q is $(v / g) \cos e c θ$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Initial velocity of particle in vector form can be written as

Question Image

$\vec{v_{P}} = v \cos θ \hat{i} + v \sin θ \hat{j} ...... (1)$
Velocity of particle at any time t will be:
$\vec{v_{Q}} = v \cos θ \hat{i} + (v \sin θ - g t) \hat{j} ..... (2)$
Given that $\vec{v_{P}} ⊥ \vec{v_{Q}}$

$∴ \vec{v_{P}}, \vec{v_{Q}} = 0$
Or $v^{2} \cos^{2} θ + v^{2} \sin^{2} θ - v \sin θ g t = 0$
Or $v^{2} = v \sin θ g t o r t = \frac{v}{g} \cos e c θ$
Substituting this value of t in Eq. (2), we get
$\vec{v_{Q}} = v \cos θ \hat{i} + (v \sin θ - \frac{v}{\sin θ}) \hat{j}$
Or $| \vec{v_{Q}} | = \sqrt{v^{2} \cos^{2} θ + v^{2} \sin^{2} θ + \frac{v^{2}}{\sin^{2} θ} - 2 v^{2}}$
$= v \cot θ$