Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A shell is projected from a level ground with a velocity of $20\;m/s$ at $45^{\circ}$ to the horizontal. When the shell is at highest point, it breaks into two equal fragments. One of the fragment whose initial velocity after the explosion is zero, falls vertically downward. At what distance from the point of projection does the other fragment fall? $(g=10\;ms^{-2})$

\[\begin {array} {1 1} (a)\;30\;m & \quad (b)\;60\;m \\ (c)\;90\;m & \quad  (d)\;40\;m \end {array}\]

Can you answer this question?

1 Answer

0 votes
Since centre of mass motion will not be affected by the explosion, it will be at a distance $R$ =range from the point of projection .
=> Total distance of the 2 nd mass from the point of projection is $\large\frac{3R}{2}$
$S= \large\frac{3}{2} \times \frac{u^2}{2g}$
$\quad= \large\frac{3}{2} \times \frac{20 \times 20}{20}$
$\quad= 30 \;m$
answered Nov 21, 2013 by meena.p
edited Jun 17, 2014 by lmohan717

Related questions

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