# Question #5c429

Jan 17, 2018

If a particle is thrown at an angle of $\theta$ w.r.t horizontal with a velocity $u$,it follows projectile motion(see the diagram) before landing.

the,upward motion is dure to the vertical component of velocity i.e $u \sin \theta$ hence it goes upward,until velocity becomes zero due to the downward acceleration due to gravity,and then comes back.

and its horizontal motion is due to the $u \cos \theta$ component.

Now,suppose it takes time $t$ to reach its highest point, so using $v = u - a t$ we get, $t = \frac{u \sin \theta}{g}$,now it will take the same time for coming down,so total time of flight becomes, $T = \frac{2 u \sin \theta}{g}$

Now,in this time $T$ if the particle horizontally goes a distance of $R$ (range) ,we can say, $R = u \cos \theta \cdot T$ (using $s = v t$ , as moving with constant velocity)

Now,putting the value of T we get,
$R = \left(u \cos \theta\right) \frac{2 u \sin \theta}{g}$ or,$\frac{{u}^{2} \sin 2 \theta}{g}$