Question

A projectile is thrown at an angle $\theta$ from the horizontal with velocity 'u' under the gravitation field of the earth. Derive expressions for its:

a)Time of its flight

b)Height

c)Horizontal Range

Answer 1

$\text{Step 1: Time of flight calculation [Refer Fig.]}$

For calculation of time of flight, we always apply equations of motion in vertical direction.

Since vertical acceleration $g$ is constant, therefore applying equation of motion in $y$ direction

Motion from O to B:

$S_y=$ Total displacement in $y=0$; $u_y=u\sin \theta ;a_y=-g$

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow$ $0=u\sin \theta t-\frac{1}{2}g{t}^2$

$\Rightarrow$ $t=\frac{2u\sin \theta }{g}$

$\text{Step 2: Horizontal range in x direction}$

$u_x=u\cos \theta ;a_x=0;s_x=R$

Since acceleration is zero in $x$ direction

So, $S_x=u_{x}t$

$\Rightarrow$ $R=u\cos \theta \times \frac{2u\sin \theta }{g}$

$\Rightarrow$ $R=\frac{u^2\sin 2\theta }{g}$

$\text{Step 3: Maximum height}$

From Figure, at maximum height only horizontal component of velocity is present, vertical component becomes zero,

So, Applying $e{q}^n$ of motion in $y$ direction between $OA$

$u_y=u\sin \theta ;v_y=0,a_y=-g$, $s_y=H$

$v_y^2-u_y^2=2a_y{s}_y$

$\Rightarrow$ $0^2-(u\sin \theta {)}^2=-2gH$

$\Rightarrow$ $H=\frac{u^2{\sin }^2\theta }{2g}$