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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A stone is thrown horizontally outward from the top of a bridge.The stone is released 19.6m above the street below.The initial velocity of the stone is 5.0m/s.Determine the magnitude and direction of the velocity of the projectile "just" before it strikes the street.

$\begin{array}{1 1}20m/s,\quad 76^{\circ}\\16 m/s, \quad 38^{\circ}\\16 m/s, \quad 76^{\circ}\\20 m/s, \quad 38^{\circ}\end{array} $

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Answer : $20 m/s, \quad 76^{\circ}$
$v_0=5.0m/s$
$y=-90.6m$
$g=9.8m/s^2$
$v_{0x}=v_0\cos\theta=5.0m/s$
$v_{0y}=v_0\sin\theta=0$
$y=v_{0y}t+\large\frac{1}{2}$$(-g)t^2$
$\Rightarrow t=2.0m/s$
$v_y=v_{0y}+(-g)t\Rightarrow v_y=-19.6m/s$
$v=\sqrt{v_x^2+v_y^2}\Rightarrow v=20.0m/s$
$\theta=\tan^{-1}\big(\large\frac{v_y}{v_x}\big)$
$\Rightarrow 76^{\large\circ}$
The direction is $76^{\large\circ}$ south of east.
answered Aug 13, 2014 by sreemathi.v
 

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