Projectile Motion
IMAT Interactive Study Tool
Two Independent Motions
Projectile motion is analyzed by splitting it into two separate components:
- Horizontal Motion: Constant velocity (zero acceleration).
- Vertical Motion: Constant downward acceleration due to gravity (\(g\)).
Initial Horizontal Velocity: \(v_{0x} = v_0 \cos(\theta)\)
Initial Vertical Velocity: \(v_{0y} = v_0 \sin(\theta)\)
Key Formulas
Time to Max Height: \(t_{peak} = \frac{v_{0y}}{g}\)
Total Time of Flight: \(T = 2 \times t_{peak} = \frac{2v_{0y}}{g}\)
Maximum Height (h): \(h = \frac{v_{0y}^2}{2g}\)
Range (R): \(R = v_{0x} \cdot T = \frac{v_0^2 \sin(2\theta)}{g}\)
Example 1: Calculating Flight Properties
A cannonball is fired with an initial velocity of 100 m/s at an angle of 30° above the horizontal. Find its time of flight and maximum height. (Use \(g = 10 \, m/s^2\), \(\sin(30^\circ)=0.5\)).
Step 1: Find the initial vertical velocity (\(v_{0y}\)).
\(v_{0y} = v_0 \sin(\theta) = 100 \times 0.5 = 50 \, m/s\)
Step 2: Calculate the total time of flight (T).
\(T = \frac{2v_{0y}}{g} = \frac{2 \times 50}{10} = 10 \, s\)
Step 3: Calculate the maximum height (h).
\(h = \frac{v_{0y}^2}{2g} = \frac{(50)^2}{2 \times 10} = \frac{2500}{20} = 125 \, m\)
Conclusion: The time of flight is 10 seconds and the maximum height is 125 meters.
These graphs show the vertical component of a projectile launched at 60° with an initial speed of 30 m/s.