Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
0 votes

Two buildings are 40m apart with what velocity should a ball be thrown horizontally from a window 145 m above the ground in one building so that it will enter a window $22.5 m$ above the ground in second building? $(g=9.8 m/s^2)$

\[(a)\;4 m/s \quad (b)\;6 m/s \quad (c)\;12 m/s \quad (d)\;8 m/s\]

Can you answer this question?

1 Answer

0 votes
Let 'u' be velocity of projection.
't' time taken for ball to fall through a vertical height of $145-22.5=122.5m$
Horizontal distance $=40 m$
$122.5=\large\frac{1}{2} $$ \times 9.8 \times t^2$
(since the vertical component of initial velocity is zero)
$t=5 \;seconds$
Also $u \times t=horizontal \;distance$
$40=u \times 5$
$u=8 m/s$
Hence d is the correct answer. 


answered Jun 29, 2013 by meena.p
edited Jan 25, 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