Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
+1 vote

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on horizontal plane is

$\begin{array}{1 1}(A)\;\large\frac{3v^2 }{g} \\(B)\;\large\frac{3}{2} \frac{u^2}{g} \\(C)\;\large\frac{ u^2}{3g}\\(D)\;\large\frac{\sqrt 3}{2}\frac{u^2}{g} \end{array} $

Can you answer this question?

1 Answer

+1 vote
As $ u \cos \theta_0 =u/2 $
$\therefore \theta_o =60^{\circ}$
So $ R= \large\frac{u^2 \sin (2 \times 60^{\circ})}{g} $
$\qquad= \large\frac{\sqrt 3}{2}\frac{u^2}{g}$
Hence D is the correct answer.
answered Jul 14, 2014 by meena.p

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