logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
0 votes

A particle is projected vertically up and it reaches a 15 cm high mark at certain instant. After 4 seconds from this instant the particle crosses this mark again in its downward journey. What is the initial velocity of projection of this particle? $(g= 10 ms^{-2})$

$\begin{array}{1 1}(A)\;10 \sqrt 7 \\(B)\;6 \sqrt 2 \\(C)\;5 \sqrt 5 \\(D)\;6.7 \sqrt 2 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
After 4 seconds , the particle returns to the same mark.
If its velocity when crossing 15 cm high mark is v then in 4s its displacement is zero.
Therefore $0= v \times 4 -\large\frac{1}{2}$$ \times 10 \times 4^2$
$v= 20 ms^{-1}$
If the initial velocity of projection is u, then
$v^2=u^2 +2as $ gives
$20^2 =u^2 -2(10)(5)$
$u^2= 700$
$u= 10 \sqrt 7 ms^{-1}$
Hence A is the correct answer.
answered Jul 25, 2014 by meena.p
 

Related questions

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