Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A projectile is given an inital velocity of (2\hat {i} + \hat{j})m/s, where \hat i is along the ground and \hat j is along the vertical. If g = 10 m/s^2 ,the equation of its trajectory is

Option: 1

4y = 2x - 25x^2


Option: 2

y = x - 5x^2


Option: 3

y = 2x - 25x^2


Option: 4

4y = 2x - 5x^2


Answers (1)

best_answer

\\ \text{Given :}

 \\ \vec{u} = (2\hat i + \hat j )m/s \\*\Rightarrow u_x = 2m/s; \;u_y = 1m/s \\ \\|\vec{u}|=\sqrt{\left ( 1 \right )^{2}+\left ( 2 \right )^{2}} =\sqrt{5} \ m/s

 \\ \text{the angle of projection}(\tan\theta) = \frac{u_y}{u_x}\\ \Rightarrow \tan\theta = \frac{1}{2}

equation of trajectory is 

y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

\\ \*\Rightarrow y = x\cdot \frac{1}{2} - \frac{10\times x^2}{2\times 5\times \frac{4}{5}} = \frac{x}{2} -\frac{5}{4}x^2 \\*\Rightarrow 4y = 2x - 5x^2

 

Posted by

Sanket Gandhi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA reopens session 2 registrations; city slip soon; admit card, exam dates
anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question