Projectile Motion and Circular Motion: Range, Trajectories, and Applications

Projectile Motion

Maximum Height of a Projectile

When a projectile is launched, its maximum height (ymax) is reached when its vertical velocity becomes zero. The following equations describe this process:

• Equation for Maximum Height: $y_{max} = \frac{v_{0y}^2}{2g}$ Where v0y is the initial vertical velocity and g is the acceleration due to gravity.

• Key Points:

  • Maximum height depends only on the initial vertical velocity and gravity.

  • Horizontal velocity does not affect maximum height.

  • Height increases quadratically with v0y and decreases with stronger gravity.

• Example: If a soccer ball is kicked upward with v0y = 11.47 m/s, then $y_{max} = \frac{(11.47)^2}{2 \times 9.8} = 6.71$ m.

Time to Reach Maximum Height

The time to reach the top of the trajectory is when the vertical velocity becomes zero:

• Equation for Time to Top: $t_{top} = \frac{v_{0y}}{g}$

• Key Points:

  • Time to top depends only on initial vertical velocity and gravity.

  • It increases linearly with v0y and decreases with stronger gravity.

• Example: For v0y = 11.47 m/s, $t_{top} = \frac{11.47}{9.8} = 1.17$ s.

Characteristics of Projectile Motion

• Constant Acceleration: The acceleration is always downward and constant (gravity).

• Horizontal Velocity: Remains constant throughout the flight.

• Vertical Velocity: Is zero at the highest point.

• Symmetry: Time spent rising equals time spent falling (to the same altitude).

• Return Speed: Returns to the original altitude at the same speed (but downward).

Range of a Projectile

The range (R) is the horizontal distance traveled by a projectile. For launch and landing at the same altitude:

• Equation for Range: $R = \frac{v_0^2 \sin 2\theta}{g}$ Where v0 is the initial speed and θ is the launch angle.

• Maximum Range: Occurs at a launch angle of 45°. $R_{max} = \frac{v_0^2}{g}$

• Example: A quarterback throws a football at 22.3 m/s at 45°: $R_{max} = \frac{(22.3)^2}{9.8} = 50.8$ m (about 55.5 yards)

Launch Angle for a Given Range

To achieve a specific range with a given speed, the launch angle can be calculated:

• Equation for Launch Angle: $\theta = \frac{1}{2} \sin^{-1}\left(\frac{gR}{v_0^2}\right)$

• Key Points:

  • Two angles can produce the same range: a low arc and a high arc.

  • For example, to throw 35 m with 20 m/s, $\theta = 29.5^\circ$ or $60.5^\circ$.

Circular Motion

Acceleration in Circular Motion

When an object moves in a circle, its acceleration is always directed toward the center (centripetal acceleration):

• Centripetal Acceleration: $a_c = \frac{v^2}{R}$ Where v is the tangential speed and R is the radius of the circle.

• Angular Velocity: $v = R\omega$ $\omega = \frac{2\pi}{T}$ Where ω is angular velocity (rad/s) and T is the period (time for one revolution).

• Key Points:

  • Acceleration is always perpendicular to velocity in uniform circular motion.

  • Results in a circular trajectory.

Example: Maximum Speed in Circular Motion

If a car can accelerate up to 8 m/s2 before sliding, and the radius of the circle is 6 m, the maximum speed is:

• Equation: $v_{max} = \sqrt{a_{max} R}$

• Calculation: $v_{max} = \sqrt{8 \times 6} = 6.93$ m/s (about 16 mph)

Summary Table: Projectile Motion Equations

Quantity	Equation	Variables
Maximum Height	$y_{max} = \frac{v_{0y}^2}{2g}$	Initial vertical velocity, gravity
Time to Top	$t_{top} = \frac{v_{0y}}{g}$	Initial vertical velocity, gravity
Range	$R = \frac{v_0^2 \sin 2\theta}{g}$	Initial speed, launch angle, gravity
Centripetal Acceleration	$a_c = \frac{v^2}{R}$	Tangential speed, radius