Question

Derive an expression for maximum height and range of an object in projectile motion.

Answer 1

Let a projectile move with a velocity $u$ which is inclined with the horizontal at angle of $\theta$. The velocity after time 't' will be

$\begin{array}{c}v=v_x\hat{i}+v_y\hat{j}\\ v=u\cos \theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j}\\ \left(Since:v_y=u\sin \theta -gt\right)\end{array}$

At maximum height the projectile will only have horizontal component that is

$\begin{array}{c}{v}_x=u\cos \theta \\ {v_y}^2-{u_y}^2=2ay\\ v_y=0\left(atmaxheightH\right)\\ u_y=u\sin \theta \\ ay=-gH\\ Puttingthesevalues,\\ 0={\left(usin\theta \right)}^2-2gH\\ H=\frac{u^2{\sin }^2\theta }{2g}\end{array}$