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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
+1 vote

A particle is projected vertically upward with a speed of $16 m/s$, after some time, when it again passes through point of projection, its speed is found to be $8 m/s$. It is known that the work done by air resistance is same during upward and downward motion, Then the maximum height attained by the body is ? $(g=10 m/s^2)$

\[(a)\;16\;m \quad (b)\;4.8 m\quad (c)\; 8 m \quad (d)\;12.8\;m\]

Can you answer this question?
 
 

1 Answer

0 votes
From work energy theorem
For upward motion
$\large\frac{1}{2}$$m(16)^2=mgh+W$----(1) (Work done against resistance)
For downward motion
$\large\frac{1}{2}$$m(8)^2=mgh-W$ -----(2)
From (1) and (2)
$\large\frac{1}{2} $$ m [(16)^2-8^2]=m2gh$
$\large\frac{1}{2}$$[256-64]=2 \times 10 \times h$
$h=4.8 \;m$
Hence b is the correct answer
answered Aug 8, 2013 by meena.p
edited Jun 13, 2014 by lmohan717
it is wrong solution. because when we add equation 1 and 2 .we should get 16 square +8 square.but you did minus.the answer is 8m not 4.8m
 

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