Question

Show that, (a) for a projectile the angle between the velocity and the x-axis as a function of time is given by $\theta \left(t\right)=ta{n}^{-1}\left(\frac{v_{o}y-gt}{v_{o}x}\right)$(b) Shows that the projection angle ${\theta }_o$ for a projectile launched from the origin is given by ${\theta }_o=ta{n}^{-1}\left(\frac{4h_m}{R}\right)$where the symbols have their usual meaning.

Answer 1

$\left(a\right)$

Consider that the initial velocity of projectile in X and Y direction be $v_{0x}$ and $v_{0y}$ respectively.

Now, as the particle reaches the point $P$ during its motion, consider $v_x$ and $v_y$ be the velocity in X and Y directions respectively.

Using the forst equation of motion for the particle to reach the point $P$.

In vertical direction:-

$v_y=v_{0y}=gt$

And in horizontal direction:=

$v_x=v_{0x}$

From the above equation, we can obtain the value of $\theta$

$tan\theta =\frac{v_y}{v_x}=\frac{(v_{0y}-gt)}{v_{0x}}$

$\theta =ta{n}^{-1}\frac{(v_{0y}-gt)}{v_{0x}}$

$\left(b\right)$

The maximum height of projectile is given by,

$h_m=\frac{u_0^{2}si{n}^2\theta }{2g}$ ...(i)

The horizontal range of the projectile is given by,

$R=\frac{u_0^{2}sin2\theta }{g}$ ... (ii)

Taking the ratio of the maximum height and range of projectile:

$\frac{h_m}{R}=\frac{sin2\theta }{2si{n}^{2}2\theta }$

$=\frac{Sin\theta \times sin\theta }{2\times 2sin\theta cos\theta }$

= $\frac{sin\theta }{4Cos\theta }=\frac{tan\theta }{4}$

$tan\theta =\left(4h_m/R\right)$

$\theta =ta{n}^{-1}\left(4h_m/R\right)$