Question

Derive the equations of projectile motion.

Answer 1

An object is thrown with a velocity $v_o$ with an angle $\theta$ with horizontal x-axis.

The velocity component along X-axis= $V_{ox}=V_{o}cos\theta$

The velocity component along Y-axis =$v_{oy}=v_{0}sin\theta$

At time T=0, no displacement along X-axis and Y-axis.

So, $x_o=0,y_0=0$

At time T=t,

Displacement along X-axis= $v_{ox}.t=v_{0}cos\theta$----- (1)

Displacement along Y-axis=$v_{oy}.t={(v}_{0}sin\theta )t-(\frac{1}{2})g{t}^2-------(2)$

Time required for maximum height,

At maximum height, the velocity component along Y-axis is zero. ie;

$v_y=0$ Let the time required to reach maximum height is $t_{max}$

Therefore, the initial velocity for motion along Y-axis

$v_y=v_{0}sin\theta -g{t}_{max}=>0=v_{0}sin\theta -g{t}_{max}=>t_{max}=\left(v_{0}sin\theta \right)/g$------- (5)

Total time of flight,

To reach maximum height, time required is $v_{o}sin\theta /g$. It takes equal time to reach back ground.

So, total time of flight, $T_{max}=\frac{2\left(v_{0}sin\theta \right)}{g}$------- (6)

Maximum height reach,]

Let the maximum height reached by the object be $H_{max}$

When body of projectile reaches the maximum height, then

$v_y^2={\left(v_{0}sin\theta \right)}^2=2g{H}_{max}=>0={\left(v_{0}sin\theta \right)}^2=2g{H}_{max}$

$\therefore H_{max}=\frac{{\left(v_{0}sin\theta \right)}^2}{2g}$---------- (7)

Horizontal range,

Let R is the horizontal range by the projected body.

Here using horizontal component of velocity only as effective velocity to transverse horizontal path.

$R=\left(v_{0}cos\theta \right).T_{max}=\frac{\left(v_{0}cos\theta \right).2\left(v_{0}sin\theta \right)}{g}=v_0^{2}sin2\theta /g$

$R=\frac{v_0^{2}sin2\theta }{g}$-------- (8)