$a)\;4 \\b)\;2 \\c)\;1 \\ d)\;0.5 $

Class11

Want to ask us a question? Click here

Browse Questions

Class11

Ad |

0 votes

0 votes

For Ball $B_2 : 0-v^2=-2gh$

$h=\large\frac{v^2}{2g}$

For Ball $B_1$ to reach the same height the vertical component of velocity is $u \sin \theta=u \sin 30$

$\quad \qquad= u/2$

Therefore $\large\frac{-u^2}{4}$$=-2g\bigg(\large\frac{v^2}{2g}\bigg)$

$\qquad=\large\frac{u^2}{v^2}$

$\qquad=4$

$\quad\large\frac{u}{v}$$=2$

Hence b is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...