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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A body is projected vertically upward with a velocity $20 ms^{-1}$ . Find the distance travelled by it in 3 s. Take $g=10ms^{-2}$.

$\begin{array}{1 1}(A)\;25\;m\\(B)\;7\;m \\(C)\;9\; m \\(D)\;67 \;m \end{array} $

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We know that a body projected vertically upward takes $\large\frac{u}{g} $ second to reach to the highest point.
Here $ u= 20 ms^{-1},a=-g=-10ms^{-2}$, therefore the body attaining maximum height, the body retraces its path.
Since the path is reversed after 2s, the relation $s= ut+ \large\frac{1}{2}$$at^2$ with $t= 3$ will not give the distance travelled.
This formula will give the displacement of the body from the ground. Thus,
$s= 20(3)-\large\frac{1}{2}$$(10)(3)^2=15 m$
The maximum height is given by $s_{max}=20(2)0-\large\frac{1}{2} $$(10)(2)^2=20 m$
Hence the distance covered by the body in 3 s is given by $20+(20-15)= 25 m$
Hence A is the correct answer.
answered Jul 25, 2014 by meena.p

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