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At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown The distance (in metres) of the particle from origin at $t = 2$ sec.
$60\sqrt 2 $
$100$
$60$
$120$
Solution
$\theta=53^{\circ}$
$u=50m/s$
$u_{x}=u \cos \theta \Rightarrow u_{x}=50 \times \frac{3}{5}$
$u_{x}=30 \mathrm{m} / \mathrm{s}$
$u_{y}=u \sin \theta \Rightarrow u_{y}50\times \frac{4}{5}$
$u_{y}=40 \mathrm{m} / \mathrm{s}$
$S=w t+\frac{1}{2} a t^{2}$
$a_{x}=0$
$a_{y}=-g \Rightarrow-10 \mathrm{m} / \mathrm{s}^{2}$
$x \rightarrow 30 \times 2+\frac{1}{2} \times 0$
$x=60 \mathrm{m}$
$y=40 x^{2}-\frac{1}{2} x+0 \times 4$
$y=80-20 \Rightarrow y=60 m$
$=\sqrt{(60)^{2}+(60)^{2}}$
$=60 \sqrt{2} \mathrm{m} / \mathrm{s}$
