Projectile motion

Acceleration and Gravity

Definition of Gravity:
- Gravity is defined as a pulling force that causes falling objects to accelerate downward to the Earth's surface.
- In the absence of any resistance from the air, the rate of change in velocity near the Earth is constant and measured as 9.8 m/s².
- This measurement is known as the acceleration due to gravity, which is often described as negative due to the downward direction of the force.
Ball's Motion During Flight:
- When a ball is tossed straight up and then caught again, its motion can be described as follows:
- At its maximum height, the ball momentarily stops (v = 0) while gravity continues to act on it, pulling it downward.
- As the ball rises, its upward velocity decreases rapidly at a rate of approximately 10 m/s every second, reflecting the acceleration due to gravity.
- On its descent, the ball's downward velocity increases at the same rate of 9.8 m/s².
- Overall, the acceleration due to gravity is always equal to 9.8 m/s² downward, being constant near the Earth's surface, regardless of the direction in which the ball is moving.

Projectile Motion in One Dimension

Definition of Projectiles:
- Projectiles are objects that move through the air and are subject to gravitational pull towards the Earth.
- Exclusions: Projectiles do not include powered vehicles like airplanes or rockets.
- Assumption: Effects of wind resistance are ignored in analysis.
Key Points for Kinematics Problems with Projectiles:
1. The acceleration due to gravity is constant at 9.8 m/s² downward near the Earth's surface.
2. In equations, the symbol for acceleration is 'a', whereas the symbol representing gravity's acceleration is 'g'. Thus, we have:
  $g = 9.8 ext{ m/s}^2$ (downward).
3. Choose a consistent direction as positive (e.g., 'up' or 'down') before solving problems.
  - If 'up' is positive, then $a = -g = -9.8 ext{ m/s}^2$ .
  - If 'down' is positive, then $a = g = 9.8 ext{ m/s}^2$ .
4. Remember that 'g' is a rate of change in velocity, similar to all applications of acceleration.

Problems with Projectile Motion in One Dimension

Example Problem 1: Golf Ball Dropped from a Height
- Scenario: A golf ball is dropped from a height of 10.0 m above the ground, neglecting wind resistance.
- Known Variables:
- Acceleration: $a = 9.8 ext{ m/s}^2$
- Distance: $d = 10.0 ext{ m}$
- Initial velocity: $V = 0$
- Find: The time $t$ it takes to hit the ground.
- Kinematics Formula Used:
  $d = vt + rac{1}{2}at^2$
- Solving Steps:
- Substitute known values: $10.0 = 0 imes t + rac{1}{2}(9.8)t^2$
  - Simplifying gives:
    $10.0 = 4.9t^2$
    $t^2 = rac{10.0}{4.9}$
    $t = ext{s } 1.43 ext{s} ext{ (rounded 1.4 s)}$
- Conclusion: The time taken to hit the ground is 1.4 s.
- Part b: The speed of the golf ball when it hits the ground will need to be calculated using further kinematic equations.
Example Problem 2: Baseball Tossed Upwards
- Scenario: A baseball is tossed straight up with an initial speed of 22.5 m/s.
- Known Variables:
- Acceleration: $a = -9.8 ext{ m/s}^2$ (because moving up)
- Initial velocity: $V = 22.5 ext{ m/s}$
- Final velocity at maximum height: $v = 0$
- Find: The maximum height $d$ the ball will reach.
- Kinematics Formula Used:
  $v^2 = V^2 + 2ad$
- Substituting Values:
- Rearranging gives: $0 = (22.5)^2 + 2(-9.8)d$
  - Simplifying leads to:
    $-506.25 = -19.6d$
    $d = rac{506.25}{19.6} <br>ightarrow 25.8 ext{m} ext{ (rounded to 26 m)}$
- Follow-up (Part b): The speed of the baseball when caught, and total time spent in the air moving up and down will also need to be calculated.

One-Dimensional Projectile Motion and the Quadratic Equation

Example Problem 3: Ball Tossed Upward with Different Heights
- Scenario: A student tosses a ball directly upwards with an initial speed of 15 m/s from ground level, intending for another student to catch it 7.8 m higher.
- Known Variables:
- Initial velocity: $V = 15 ext{ m/s}$
- Overall displacement: $d = 7.8 ext{ m}$
- Acceleration: $a = -9.8 ext{ m/s}^2$
- Time: $t = ?$
- Using Kinematic Equation:
- Substitute all into: $d = Vt + rac{1}{2}at^2$ $7.8 = 15t + rac{1}{2}(-9.8)t^2$
  - Rearranging gives:
    $-4.9t^2 + 15t - 7.8 = 0$
- Solving the Quadratic Equation:**
- Using the quadratic formula: $t = rac{-b \pm ext{ } ext{sqrt}(b^2 - 4ac)}{2a}$
  - Inserting values:
    $t = rac{-15 \pm ext{ } ext{sqrt}(15^2 - 4(-4.9)(-7.8))}{2(-4.9)}$
- Determining Final Velocity and Time:**
- Find the ball's final velocity upon being caught, correlating that with the air-time confirmed equal to 2.4 s.

Graphing Analysis of One-Dimensional Projectile Motion

Distance-Time Graph Analysis:
- A graph illustrates the motion of a baseball tossed up and then caught at the same position:
- The maximum height is reached at the halfway point in time.
- The slope of the distance-time graph reflects the velocity, indicating initial motion with a steep positive slope.
- As time progresses, the velocity decreases, as shown by a decreasing slope.
- At the maximum height, the velocity becomes zero, indicating a horizontal slope.
- Post maximum height, the velocity accelerates downwards, illustrated as an increasing negative slope.
- The slopes during ascent and descent are equal and opposite.
Velocity-Time Graph Analysis:
- A graph for the velocity of a baseball tossed up and then caught:
- Initial positive velocity starts at the y-intercept and decreases to zero at the peak height.
- It then increases in the negative direction as the ball falls back.
- The straight line slope indicates a constant acceleration of -9.8 m/s².
- The area under the graph yields the ball's displacement—indicating a positive area during the rise and an equal negative area during the fall.