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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A spring gun placed on a table fires a steal ball at a $45^{\large\circ}$ angle above the horizontal.The ball leaves the muzzle of the gun 1.1m above the floor and travels 4.6m horizontally.Determine the total time that the ball is in the air.

$\begin{array}{1 1}1.08s\\0.54s\\2.16s\\0.90s\end{array} $

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Answer : 1.08s
$\theta=45^{\large\circ}$
$y=-1.1m$
$v_{0x}=v_0\cos\theta=\large\frac{\sqrt 2}{2}$$v_0$
$v_{0y}=v_0\sin\theta=\large\frac{\sqrt 2}{2}$$v_0$
$y=v_{0y}t+\large\frac{1}{2}$$(-g)t^2$
$x=v_{0x}t$
$\Rightarrow -1.1=\large\frac{\sqrt 2}{2}$$v_0t-4.9t^2$
$\Rightarrow 4.6=\large\frac{\sqrt 2}{2}$$v_0t$
$\Rightarrow t=1.08s$
answered Aug 13, 2014 by sreemathi.v
 

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