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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A projectile is fired with an initial speed of 113m/s at an angle of $60.0^{\large\circ}$ above the horizontal from the top of a cliff 49.0m high.Determine the tome to reach maximum height,maximum height above the base of the cliff reached by the projectile.

$\begin{array}{1 1}t = 10s, \quad h=538 m\\t = 5s, \quad h=538 m\\t = 5s, \quad h=489m\\t = 10s, \quad h=489 m\end{array} $

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Answer : $t = 10s, \quad h=538 m$
$v_0=113m/s$
$g=9.8m/s^2$
$v_{0x}=v_0\cos\theta=56.5m/s$
$v_{0y}=v_0\sin \theta=97.9m/s$
$v_y=v_{0y}+(-g)t$
$\Rightarrow t=10.0s$
$y=v_{0y}t+\large\frac{1}{2}$$(-g)t^2$
$\Rightarrow y=489m$
$h=489m+49m$
$\Rightarrow h=538m$
answered Aug 13, 2014 by sreemathi.v
 

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