# Applications of Projectile Motion

## Example

A particle is projected at an angle of $45^{\circ}$ and a velocity of $60 \text{ ms}^{-1}$, and taking gravity as $10 \text{ ms}^{-2}$.

(a)     Find the equation for the particle for horizontal displacement.

\begin{align} \displaystyle \require{AMSsymbols} \ddot{x} &= 0 \\ \dot{x} &= 60 \cos 45^{\circ} = 60 \times \frac{1}{\sqrt{2}} = 30 \sqrt{2} \\ x &= \int{\dot{x}} dt = \int {30 \sqrt{2}} dt = 30 \sqrt{2} t + C \\ 0 &= 30 \sqrt{2} \times 0 + C &\color{green}{\text{initially } t=0 \text{ and } x=0} \\ C &= 0 \\ \therefore x &= 30 \sqrt{2} t \end{align}

(b)     Find the initial displacement of the vertical motion.

$\displaystyle \dot{y} = 60 \sin 45^{\circ} = 60 \times \frac{1}{\sqrt{2}} = 30 \sqrt{2}$

(c)     Find the equation for the particle for vertical displacement.

\begin{align} \displaystyle \require{AMSsymbols} \ddot{y} &= -10 \\ \dot{y} &= \int {\ddot{y}} dt = \int {-10} dt = -10t + C \\ 30 \sqrt{2} &= -10 \times 0 + C \\ C &= 30 \sqrt{2} \\ \dot{y} &= -10 t + 30 \sqrt{2} \\ y &= \int {-10t+30 \sqrt{2}} dt \\ &= -5t^2 + 30 \sqrt{2} t + D \\ 0 &= -5 \times 0^2 + 30 \sqrt{2} \times 0 + D &\color{green}{t=0 \text{ and } y=0} \\ D &= 0 \\ \therefore y &= -5t^2 + 30 \sqrt{2} t \end{align}

(d)     Find the time taken to reach the ground.

\begin{align} \displaystyle \require{AMSsymbols} y &= 0 &\color{green}{\text{to reach the ground}} \\ -5t^2 + 30 \sqrt{2} t &= 0 \\ 5t (-t+6\sqrt{2}) &= 0 \\ t &= 0 \text{ or } -t+6\sqrt{2} = 0 \\ \therefore t &= 6 \sqrt{2} \text{ seconds} \end{align}

(e)     Find the horizontal distance that the particle travels.

\begin{align} \displaystyle \require{AMSsymbols} \text{horizontal distance at } t = 6 \sqrt{2} \\ \text{substituting } t = 6 \sqrt{2} \text{ into } x=30 \sqrt{2} t \\ x = 30 \sqrt{2} \times 6 \sqrt{2} = 360 \text{ metres} \end{align}

(f)     Find the maximum height of the particle.

\begin{align} \displaystyle \require{AMSsymbols} \text{maximum height occurs at } \dot{y} &= 0 \\ -10 t + 30 \sqrt{2} &= 0 \\ y_{\text{max}} &= -5 \left(3\sqrt{2} \right)^2 + 30 \sqrt{2} \times 3 \sqrt{2} \\ &= 90 \text{ m} \end{align}

(g)     Find the equation of the trajectory.

\begin{align} \displaystyle \require{AMSsymbols} t &= \frac{x}{30\sqrt{2}} &\color{green}{\text{equation of trajectory is obtained by eliminating } t} \\ y &= -5 \times \left( \frac{x}{30\sqrt{2}} \right)^2 + 30 \sqrt{2} \times \frac{2}{30\sqrt{2}} \\ \therefore y &= -\frac{x^2}{360} + x \end{align}

(h)     Find the height of the particle when the horizontal distance is $10$ m.

\begin{align} \displaystyle \require{AMSsymbols} \text{substitute } x &= 10 \text{ into } y = -\frac{x^2}{360} + x \\ y &= -\frac{10^2}{360} + 10 = 9.72 \text{ m} \end{align}

(i)     Find the horizontal distance when the height of the particle is $50$ m.

\begin{align} \displaystyle \require{AMSsymbols} \text{substitute } y &= 50 \text{ into } y = -\frac{x^2}{360} + x \\ -\frac{x^2}{360} + x = 50 \\ x^2-360x+18000 &= 0 \\ (x-60)(x-300) &= 0 \\ \therefore x &= 60 \text{ and } 300 \text{ m} \end{align}