Question

A) Show that for a projectile the angle between the velocity and the 𝑥-axis as a function of time is given by, $\theta \left(t\right)={\tan }^{-1}\left(\frac{v_{0y}-gt}{v_{0x}}\right)$ where the symbols have their usual meaning.

B) Show that the projection angle ${\theta }_0$ for a projectile launched from the origin is given by ${\theta }_0={\tan }^{-1}\left(\frac{4h_m}{R}\right)$ where the symbols have their usual meaning.

Answer 1

A) Let component of $v_0$ along x–axis $=v_{0x}$
Component of $v_0$ along 𝑦–axis $=v_{0y}$
Let $v_x$ and $v_y$ respectively be the horizontal and vertical component of velocity at a point P.

Time taken by the projectile to reach point P = t
Since there is no acceleration in the horizontal direction, we can write the component of velocity ‘v’ at any instant ‘t’, as
$v_x=v_{0x}=v_0\cos {\theta }_0=\text{constant}$
$v_y=v_{0y}-gt=v_0\sin {\theta }_0-gt$
$\tan \theta \left(t\right)=\frac{v_y}{v_x}=\frac{v_{0y}-gt}{v_{0x}}$
$\theta \left(t\right)={\tan }^{-1}\left(\frac{v_{0y}-gt}{v_{0x}}\right)$

B) We assume the maximum height of projectile is $h_m$.
So, $h_m=\frac{v_0^2{\sin }^2{\theta }_0}{2g}.....\left(1\right)$
and horizontal range,
$R=\frac{v_0^2\sin 2{\theta }_0}{g}....\left(2\right)$
Dividing equation (1) by equation (2), we get
$\frac{h_m}{R}=\frac{v_0^2{\sin }^2{\theta }_0}{2g}\times \frac{g}{v_0^2\sin 2{\theta }_0}=\frac{{\sin }^2{\theta }_0}{2\times 2\sin {\theta }_0\cos {\theta }_0}$
$=\frac{1}{4}\tan {\theta }_0$
${\theta }_0={\tan }^{-1}\left(\frac{4h_m}{R}\right)$