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2.Motion in Straight Line
easy
A body is thrown vertically upwards with a velocity of $19.6\, ms^{-1}$. The position of the body after $4\, s $ will be
A
at the highest point
B
at the mid-point of the line joining the starting point and the highest point
C
at the starting point
D
none of the above
(AIIMS-2009)
Solution
$\begin{array}{l}
Clearly\,the\,time\,taken\,by\,the\,particle\\
to\,reach\,the\,highest\,{\rm{ point}}\,is\,given\,by\\
V = u – gt\\
or,\,t = \frac{{u – v}}{g} = \frac{{19.6 – 0}}{{9.8}}\\
or,\,t = 2\,s.\\
Therefore,\,the\,particle\,will\,reach\,at\,the\\
starting\,{\rm{point}}\,{\rm{itself}}\,{\rm{after}}\,{\rm{4}}\,s.
\end{array}$
Standard 11
Physics
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