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(t) = tan^-1 (voy - gt/v0x)

(b) show that the projection angle Theta_0 for a projectile launched from the origin is given by theta 0 = tan^-1(4h_m/R)​

Answer 1

Aηswer:

(a) Let $\displaystyle\sf v_{0_{x}}$ and $\displaystyle\sf v_{0_{y}}$ respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.

Let $\displaystyle\sf v_x$ and $\displaystyle\sf v_y$ respectively be the horizontal and vertical components of velocity at a point P.

time taken by projectile to reach point P = t.

• Applying the 1st equation of motion along the vertical and horizontal directions, we get

$\displaystyle\sf v_y = v_{0_{y}} = gt$

and $\displaystyle\sf v_x = v_{0_{x}}$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\;\:\;\: \therefore \tan\theta = \dfrac{v_y}{v_x} = \dfrac{v_{0_{y}} - gt}{v_{0_{x}}}$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\;\:\;\: \boxed{\sf\theta = \tan^{-1} \left(\dfrac{v_{0_{y}} - gt}{v_{0_{x}}}\right)}$

$\displaystyle\sf$

(b) Maximum vertical height

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\;\:\;\: h_m = \dfrac{x^2_0 \sin^2\theta}{2g}\:\:\;\dots(i)$

$\displaystyle\sf$

Horizontal range,

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\;\:\;\:R = \dfrac{x^2_0\sin^22\theta}{g}\:\;\;\dots(ii)$

Solving eq. (i) & (ii),

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\;\:\;\:\dfrac{-h_m}{R} = \dfrac{\sin^2\theta}{2\sin^2\theta}$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\:\;\;\;\:\;\;\:\;\: = \dfrac{\sin\theta\times\sin\theta}{2\times2\sin\theta\cos\theta}$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\:\;\;\;\:\;\;\:\;\: = \dfrac{1}{4}\:\dfrac{\sin\theta}{\cos\theta}$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\:\;\;\;\:\;\;\:\;\: = \dfrac{1}{4} \:\tan\theta$

$\displaystyle\sf$

$\displaystyle\sf \:\:\:\;\;\;\;\;\;\:\;\;\;\:\;\;\:\;\:\tan \theta = \left(\dfrac{4h_m}{R}\right)$

$\displaystyle\sf$

$\displaystyle\sf\:\:\:\;\;\;\;\;\;\:\;\;\;\:\;\;\:\;\: \boxed{\sf \theta = \tan^{-1}\left(\dfrac{4h_m}{R}\right)}$