Answer to Question #169963 in Calculus for Vishal

Question

A Cannon ball is shot from the ground with velocity 'v' at an angle 'a' , so that it has a vertical

component of velocity 'vsina' and a horizontal Component 'vcosa' . Its distance s(t) above the ground obeys the law s(t) = - 16t² + (vsina)t , while its horizontal velocity remains constantly

'vcosa'

(a) show that the path of the Cannon ball is a parabola ( find the position at each time t

and show that these points lie on parabola)

(b) find the angle 'a' which will maximize the horizontal distance traveled by the Cannon ball

before striking the ground.

Answer 1

a) Horizontal coordinate we can found from horizontal speed:

$x(t) = \int v_xdt = \int vcosa dt = vcosa \cdot t + x_0$

$=> t = (x - x_0) / (vcosa)$

Next we substitute it into the s(t) formula (let y(t) = s(t)):

$y(t) = -16t^2 + vsina\cdot t = -16\cdot (x - x_0)^2/(vcosa)^2 + vsina\cdot (x-x_0) / (vcosa) = -16\cdot (x - x_0)^2/(vcosa)^2 + tga\cdot (x-x_0)$

This is equation of parabola which connects vertical and horizontal coordinates. That is why the path of the Cannon ball is a parabola.

b) The time needed for Cannon ball to reach the highest point of path:

$v_y = s'(t) = -32t_p + vsina = 0 => t_p = vsina / 32$

Then length of horizontal path is:

$L = v_x \cdot 2t_p = vsina \cdot 2sina/32 = vsin2a / 32$

This function of parameter alpha reach maximum when $a = \frac{\pi}{4}$