Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
0 votes

The trajectory of a projectile in a vertical plane is $y=ax-bx^2$ where $a$ and $b$ are constants. The maximum height attained by the particle and the angle of projection from horizontal are

\[(a)\;\frac{b^2}{2a}, \tan ^{-1} (b)\quad (b)\;\frac{a^2}{4b},\tan ^{-1}(a) \quad (c)\;\frac{a^2}{b},\tan ^{-1} (2a) \quad (d)\;\frac{2a^2}{b},\tan ^{-1}(a)\]

Can you answer this question?

2 Answers

0 votes
for maximum height
Therefore $y_{\large max}=a \bigg(\large\frac{a}{2b}\bigg)$$-b\bigg(\large\frac{a}{2b}\bigg)^2=\frac{a^2}{4b}$
angle of projection $\tan {\theta}=> at \;x=0$
$\quad\qquad\qquad=a-2bx \bigg|_{x=0}$
Therfore $\tan \theta =a$
$\qquad \theta=\tan ^{-1} (a)$
Hence b is the correct answer. 


answered Jul 1, 2013 by meena.p
edited Jan 26, 2014 by meena.p
0 votes
answer is (B)

a^2/4b, tan^-1
answered Apr 24 by vipul.saini442

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App