Projectile Motion: MDCAT Physics notes
Projectile Motion notes for MDCAT: parabolic path, independent horizontal and vertical motion, velocity and acceleration at the top, and common traps.
What is projectile motion?
A projectile is any body thrown into the air that then moves under gravity alone (air resistance neglected). Its motion is two-dimensional motion under constant acceleration due to gravity, taking place in a vertical plane. Examples: a stone thrown horizontally from a roof, a football kicked at an angle, a bullet fired from a gun.
Independence of the two components
The key idea is that the horizontal and vertical motions are independent of each other. Gravity acts only vertically, so it changes only the vertical component of velocity.
| Quantity | Horizontal (x) | Vertical (y) |
|---|---|---|
| Force | Zero | Weight $mg$ downward |
| Acceleration | $a_x = 0$ | $a_y = -g$ (constant) |
| Initial velocity | $v_{ix} = v_i\cos\theta$ | $v_{iy} = v_i\sin\theta$ |
| Velocity at time t | $v_x = v_i\cos\theta$ (constant) | $v_y = v_i\sin\theta - gt$ |
| Displacement | $x = v_i\cos\theta\, t$ | $y = v_i\sin\theta\, t - \tfrac12 g t^2$ |
Because $x$ grows linearly with time while $y$ contains a $t^2$ term, eliminating $t$ gives $y$ as a quadratic in $x$. The trajectory is therefore a parabola. This is true whether the body is thrown at an angle or horizontally from a height.
Velocity along the path
- At every point, speed $v = \sqrt{v_x^2 + v_y^2}$ and direction $\tan\phi = v_y/v_x$.
- At the highest point: $v_y = 0$, but $v_x = v_i\cos\theta$ remains. So the velocity is minimum but not zero and is horizontal.
- Velocity is maximum at the point of projection (and equal in magnitude at landing on the same level).
- Acceleration is $g$ downward at every point, including the top. It is never zero during flight.
Horizontal launch from a height
For a body thrown horizontally with speed $u$: $v_{iy} = 0$. After time $t$, horizontal velocity is still $u$, vertical velocity is $gt$ downward. Example: a ball kicked horizontally at $15\ \text{m s}^{-1}$; after 2 s (taking $g = 10\ \text{m s}^{-2}$), $v_x = 15\ \text{m s}^{-1}$ and $v_y = 20\ \text{m s}^{-1}$ downward, so the speed is $25\ \text{m s}^{-1}$.
Role of mass
All bodies fall with the same acceleration $g$, so the mass has no effect on the path, time of flight, height or range. Only the launch speed, launch angle and $g$ matter.
Key formulas
- $v_x = v_i\cos\theta$, $\quad v_y = v_i\sin\theta - gt$
- $x = v_i\cos\theta\,t$, $\quad y = v_i\sin\theta\,t - \tfrac12 gt^2$
- Velocity at the top: $v_{top} = v_i\cos\theta$
- Resultant speed: $v = \sqrt{v_x^2+v_y^2}$
Common MDCAT traps
- Velocity at the highest point is not zero; only the vertical component is zero. It equals $v_i\cos\theta$.
- Acceleration at the peak is still $g$, not zero.
- Horizontal velocity does not change; there is no horizontal acceleration or horizontal force.
- Vertical acceleration is constant; vertical velocity is not.
- Mass does not affect projectile motion; do not choose "mass" as a factor.
Quick revision
- Path of a projectile: parabola.
- Horizontal and vertical motions are independent.
- $a_x = 0$, $a_y = g$ downward throughout.
- Speed is least at the top ($v_i\cos\theta$) and greatest at launch.
- Horizontal throw: $v_x$ stays constant, $v_y = gt$.