Section 1.4: Projectile Motion
This section introduces projectile motion as a special case of 2D motion under constant acceleration (gravity). We break motion into horizontal and vertical components and analyze trajectories, range, and maximum height.
No acceleration (ignoring air resistance): \( x = v_x t \)
Velocity: \( v_x = \text{constant} \)
Acceleration due to gravity: \( a_y = -g \)
Equations: \( y = v_{y0} t - \frac{1}{2} g t^2 \), \( v_y = v_{y0} - g t \)
Path of projectile: \( y(x) = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta} \)
Range: \( R = \frac{v_0^2 \sin 2\theta}{g} \)
Maximum height: \( H = \frac{v_0^2 \sin^2 \theta}{2g} \)
Example 1
A projectile is launched at \(v_0 = 20\text{ m/s}\) at \(30^\circ\) above the horizontal. Find maximum height and range.
Vertical component: \( v_{y0} = 20 \sin30^\circ = 10\text{ m/s} \)
Maximum height: \( H = \frac{v_{y0}^2}{2g} = \frac{10^2}{2*9.8} \approx 5.10\text{ m} \)
Horizontal component: \( v_x = 20 \cos30^\circ \approx 17.32\text{ m/s} \)
Time of flight: \( t_f = \frac{2 v_{y0}}{g} = \frac{2*10}{9.8} \approx 2.04\text{ s} \)
Range: \( R = v_x t_f \approx 17.32 * 2.04 \approx 35.3\text{ m} \)
Practice Problems
- A projectile is launched at 15 m/s at 45°. Find max height and range.
- A ball is thrown horizontally from 20 m high with 10 m/s. How far does it land?
- A projectile has range 50 m and initial speed 20 m/s. Find launch angle.
- An object is thrown at 25 m/s at 60°. Find time of flight.
- A projectile launched at 12 m/s reaches 5 m max height. Find vertical component of velocity.