Introduction to Projectile Motion
Projectile motion is a form of motion experienced by an object or particle that is projected in a gravitational field, such as a ball thrown into the air. By ignoring air resistance, we can treat the horizontal and vertical motions as independent.
1. Showing the Trajectory is a Parabola
Let an object be projected with velocity $u$ at an angle $\theta$ to the horizontal.
- Horizontal component: $u_x = u \cos \theta$, acceleration $a_x = 0$.
- Vertical component: $u_y = u \sin \theta$, acceleration $a_y = -g$.
At any time $t$, the coordinates $(x, y)$ are:
- $x = (u \cos \theta)t \implies t = \frac{x}{u \cos \theta}$
- $y = (u \sin \theta)t - \frac{1}{2}gt^2$
Substituting $t$ into the $y$ equation: $y = (u \sin \theta)(\frac{x}{u \cos \theta}) - \frac{1}{2}g(\frac{x}{u \cos \theta})^2$ $y = x \tan \theta - (\frac{g}{2u^2 \cos^2 \theta})x^2$
This is of the form $y = Ax - Bx^2$, which is the equation of a parabola.
2. Deriving Time of Flight ($T$)
The total time of flight is when the projectile returns to the same vertical level ($y=0$). $0 = (u \sin \theta)T - \frac{1}{2}gT^2$ $T(u \sin \theta - \frac{1}{2}gT) = 0$ Since $T \neq 0$, we get $T = \frac{2u \sin \theta}{g}$.
3. Deriving Maximum Height ($H$)
At maximum height, the vertical velocity $v_y = 0$. Using $v_y^2 = u_y^2 + 2a_y s_y$: $0 = (u \sin \theta)^2 - 2gH$ $2gH = u^2 \sin^2 \theta$ $H = \frac{u^2 \sin^2 \theta}{2g}$