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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
+1 vote

A projectile is thrown into space so as to have the maximum possible horizontal range =400m.Taking the point of projection as the origin,the co-ordinate of the point where the velocity of the projectile is minimum,are

$\begin{array}{1 1}400m,100m\\200m,100m\\400m,200m\\200m,200m\end{array} $

Can you answer this question?
 
 

1 Answer

+1 vote
The velocity will be minimum at the highest point.
So,x-co-ordinate is $\large\frac{400}{2}$ m ,i.e.,200m
Again,$y=h_{max}=\large\frac{v^2\sin^245^{\large\circ}}{2g}$
$\Rightarrow \large\frac{v^2}{4g}$
$\Rightarrow \large\frac{1}{4}$$\times 400m=100m$
answered Aug 13, 2014 by sreemathi.v
 

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