A particle is projected from horizontal XZ plane with velocity (uxi^+uyj^+yzk^) from origin (+y axis is upward). Find time(s) when velocity of particle will be inclined at an angle of 530 with horizontal.

3 u y − 4 u x 2 + u z 2 3 g , 3 u y + 4 u x 2 + u z 2 3 g

tan 37 0 = V y u x 2 + u z 2 Here v y = u y − g t

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

A particle is projected from horizontal XZ plane with velocity $(u_{x} \hat{i} + u_{y} \hat{j} + y_{z} \hat{k})$ from origin $(+ y axis is upward)$ . Find time(s) when velocity of particle will be inclined at an angle of $53^{0}$ with horizontal.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{u_{y} - \sqrt{u_{x}^{2} + u_{z}^{2}}}{g}$

$\frac{3 u_{y} - 4 \sqrt{u_{x}^{2} + u_{z}^{2}}}{3 g}$

$\frac{3 u_{y} + 4 \sqrt{u_{x}^{2} + u_{z}^{2}}}{3 g}$

$\frac{u_{y} + \sqrt{u_{x}^{2} + u_{z}^{2}}}{3 g}$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

$\tan 37^{0} = \frac{V_{y}}{\sqrt{u_{x}^{2} + u_{z}^{2}}}$
Here $v_{y} = u_{y} - g t$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE