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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 total time it stays in the air,and horizontal range of the projectile.

$\begin{array}{1 1}t = 20.5s, \quad \text{range} = 1156m\\t = 20.5s, \quad \text{range} = 97.9 m\\t = 10.25s, \quad \text{range} = 56.5 m\\t = 10.25s, \quad \text{range} = 1032m\end{array} $

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Answer : $t = 20.5s, \quad \text{range} = 1156m$
$v_0=113m/s$
$\theta=60.0^{\large\circ}$
$v_{0x}=v_0\cos\theta=56.5m/s$
$v_{0y}=v_0\sin\theta=97.9m/s$
$y=-49m$
$y=v_0t+\large\frac{1}{2}$$(-g)t^2$
$\Rightarrow t=20.5s$
$x=v_xt$
$\Rightarrow x=1156m$
answered Aug 13, 2014 by sreemathi.v
 

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