how to find the parabola of a flying object

Question

how can you find the parabola of a flying object without testing it? what variables do you need? I want to calculate the maximum hight and distance using a parabola. Is this possible? Any help will be appreciated.

Answer 1

I assume that you know that if an object is thrown straight upwards, with initial speed $v$, then its height $h(t)$ above the ground at time $t$ is given by $$h(t)=vt-\frac{1}{2}gt^2,$$ where $g$ is the acceleration due to gravity. The acceleration is taken to be a positive number, constant since if our thrown object achieves only modest heights. In metric units, $g$ at the surface of the Earth is about $9.8$ metres per second per second. Of course the equation only holds until the object hits the ground. We are assuming that there is no air resistance, which is unrealistic unless we are on the Moon.

Now imagine that we are standing at the origin, and throw a ball with speed $s$, at an angle $\theta$ to the ground, where $\theta$ is not $90$ degrees.

The horizontal component of the velocity is $s\cos\theta$, and the vertical component is $s\sin\theta$. So the "$x$-coordinate" of the position at time $t$ is given by $$x=x(t)=(s\cos\theta)t.\qquad\qquad(\ast)$$ The height ($y$-coordinate) at time $t$ is given by
$$y=y(t)=(s\sin\theta)t-\frac{1}{2}gt^2.\qquad\qquad(\ast\ast)$$ To obtain the equation of the curve described by the ball, we use $(\ast)$ to eliminate $t$ in $(\ast\ast)$.

From $(\ast)$ we obtain $t=\dfrac{x}{s\cos\theta}$. Substitute for $t$ in $(\ast\ast)$. We get $$y=(\tan\theta)x-\frac{g}{2s^2\cos^2\theta}x^2.$$

For the maximum height reached, we do not need the equation of the parabola. For the height $y$ at time $t$ is $(s\sin\theta)t -\frac{1}{2}gt^2$. We can find the $t$ that maximizes height. More simply, the vertical component of the velocity at time $t$ is $s\sin\theta -gt$, and at the maximum height this upwards component is $0$. Now we can solve for $t$.

Nor do we need the equation of the parabola to find the horizontal distance travelled. By symmetry, to reach the ground takes time equal to twice the time to reach maximum height. The time to reach the ground is therefore $\dfrac{2s \sin\theta}{g}$, and therefore the horizontal distance travelled is $\dfrac{2s^2\sin\theta\cos\theta}{g}$. This can also be written as $\dfrac{s^2\sin 2\theta}{g}$.

Comment: Note that for fixed $s$ the horizontal distance $\dfrac{s^2\sin 2\theta}{g}$ travelled until we hit the ground is a maximum when $\theta$ is $45$ degrees. So for maximizing horizontal distance, that's the best angle to throw at, if your throw speed does not depend on angle. (But it does!)