Vertical Motion Under Constant Acceleration: Kinematics of Free Fall

Motion Along a Straight Line

Vertical Motion Under Gravity

Vertical motion under gravity is a classic example of motion along a straight line with constant acceleration. In these problems, the acceleration is due to gravity, and the motion is typically analyzed using kinematic equations.

• Displacement (y): The change in position of the object along the vertical axis.

• Velocity (v): The rate of change of position. It can be positive (upward) or negative (downward) depending on the chosen reference direction.

• Acceleration (a): For free fall near Earth's surface, acceleration is constant and equal to $a = -g$ (where $g \approx 9.8\ \mathrm{m/s^2}$ downward).

• Initial Conditions: The initial position ($y_0$) and initial velocity ($v_0$) are crucial for solving kinematic problems.

Kinematic Equations for Constant Acceleration

These equations describe the motion of an object under constant acceleration:

• Displacement as a function of time: $y = y_0 + v_0 t + \frac{1}{2} a t^2$

• Velocity as a function of time: $v = v_0 + a t$

• Velocity as a function of displacement: $v^2 = v_0^2 + 2a(y - y_0)$

Choosing the Reference Frame

• The direction of the positive axis (up or down) must be chosen and used consistently.

• Acceleration due to gravity is negative if up is positive, and positive if down is positive.

• Initial position ($y_0$) is often set to zero for convenience.

Applications: Free Fall Problems

Case 1: Object Released from Rest

When an object is dropped from rest ($v_0 = 0$), the equations simplify:

• Displacement: $y = y_0 + \frac{1}{2} a t^2$

• Velocity: $v = a t$

• Example: Dropping a ball from a height $h$; find the time to hit the ground:

Set $y_0 = h$, $y = 0$, $a = -g$:

• $0 = h + 0 \cdot t - \frac{1}{2} g t^2$

• $t = \sqrt{\frac{2h}{g}}$

Case 2: Object Thrown Upward

If an object is projected upward with initial velocity $v_0$:

• At maximum height: $v = 0$

• Time to reach maximum height: $0 = v_0 - g t_{max}$ $t_{max} = \frac{v_0}{g}$

• Maximum height reached: $y_{max} = y_0 + v_0 t_{max} - \frac{1}{2} g t_{max}^2$ Substitute $t_{max}$: $y_{max} = y_0 + \frac{v_0^2}{2g}$

• Total time in air (flight time): For an object thrown upward and returning to the same level: $t_{flight} = 2 t_{max} = \frac{2 v_0}{g}$

• Final velocity upon return: $v = -v_0$ (same magnitude, opposite direction)

Worked Example

• Given: $v_0 = 15\ \mathrm{m/s}$ upward, $y_0 = 0$

• Find: (a) Time to reach maximum height, (b) Maximum height, (c) Total flight time, (d) Final velocity upon return to ground.

1. Time to maximum height: $t_{max} = \frac{v_0}{g} = \frac{15}{9.8} \approx 1.53\ \mathrm{s}$

2. Maximum height: $y_{max} = \frac{v_0^2}{2g} = \frac{15^2}{2 \times 9.8} \approx 11.5\ \mathrm{m}$

3. Total flight time: $t_{flight} = 2 t_{max} = 3.06\ \mathrm{s}$

4. Final velocity: $v = -v_0 = -15\ \mathrm{m/s}$ (downward)

Quadratic Equation in Kinematics

When solving for time or displacement, the kinematic equation may reduce to a quadratic form:

• $y = y_0 + v_0 t + \frac{1}{2} a t^2$

• Rearranged: $0 = \frac{1}{2} a t^2 + v_0 t + (y_0 - y)$

• Solution for $t$ (using quadratic formula): $t = \frac{-v_0 \pm \sqrt{v_0^2 - 2a(y_0 - y)}}{a}$

Note: The quadratic formula is useful for finding the time when the object reaches a certain position.

Key Concepts and Summary Table

Quantity	At Launch	At Maximum Height	At Return to Ground
Position ($y$)	$y_0$	$y_{max}$	$y_0$
Velocity ($v$)	$v_0$	$0$	$-v_0$
Time ($t$)	$0$	$t_{max}$	$t_{flight}$

Additional info:

• Air resistance is neglected in these calculations.

• All equations assume constant acceleration (i.e., gravity is uniform).

• These principles apply to any straight-line motion with constant acceleration, not just vertical motion.