Projectile Motion Lecture Notes

Projectile Motion

Definition

• Projectile motion is a form of motion (kinematics) where an object is thrown near the Earth's surface, and its motion is only affected by gravity.

Gravity

• Gravity constantly pulls objects towards the Earth with a constant acceleration.
• The acceleration due to gravity is approximately $9.8 \frac{m}{s^2}$.

Simulation

• The PhET simulation demonstrates the force of gravity and the behavior of various masses under the influence of gravity.

Example: Buick Shot from a Cannon

• Initial velocity: $18 \frac{m}{s}$ (directly upwards).
• Mass: $1000 kg$.
• The Buick goes up and then comes down, following a parabolic path.
• The shape of the position of the Buick over time looks like a parabola, indicating constant acceleration due to gravity.

Key Observations

• Increasing the initial velocity of the Buick causes it to fly higher.
• The velocity of the Buick starts as a high positive number, slows down to zero, and then becomes a high negative number as it speeds towards the ground.
• Position has a parabolic shape, velocity is a straight line with a non-zero slope, and acceleration is a non-zero straight line with zero slope.

Mass and Projectile Motion

• Changing the mass does not significantly alter the motion of an object if there is no air resistance or friction.
• A baseball will have the same flight pattern as a Buick in a vacuum.
• Kinematic expressions from lecture 2.1 do not depend on mass.
• The final velocity of a Buick is the same as that of a baseball when shot upwards.

Air Resistance (Drag)

• Air resistance affects objects with less mass more significantly.
• The Buick, with a large mass, is not easily affected by air resistance.
• The baseball, with a small mass, is greatly affected by air resistance.

Projectile Motion on a Cliff

• Throwing a rock straight up or dropping it off a cliff can both be modeled as projectile motion.
• The position, velocity, and acceleration graphs are similar to other constant acceleration situations.

Two-Dimensional Projectile Motion

• When the Buick is shot at an angle of zero degrees with respect to the horizontal, it drops a small distance in the vertical direction and moves a little bit in the horizontal direction.
• Vectors have two dimensions, so we need to consider horizontal and vertical components independently.
• At a 45-degree angle, the horizontal distance traveled is much greater.
• Vertical velocity is positive, then zero, and then negative, indicating constant acceleration downwards due to gravity.
• Horizontal velocity is constant.
• Horizontal position increases linearly with non-zero slope, indicating constant velocity.
• Hang time connects the two dimensions; the time of flight is dependent on the time the projectile is allowed to accelerate in the vertical direction before hitting the ground.
• Projectile motion experiences constant acceleration in the vertical direction due to gravity and constant velocity in the horizontal direction.

Problem Solving

• Use constant velocity and constant acceleration concepts to solve for unknown variables.
• Visualize the behavior of a projectile from everyday experiences.

Example: Basketball

• Initial velocity vectors have horizontal and vertical components.
• The vertical component experiences acceleration and is constantly changing.
• The horizontal component remains the same until the projectile hits the ground.
• There is symmetry in the vertical component of velocity: the velocity at each instant before the peak is the same magnitude as the velocity at each instant after the peak, but in the negative direction.

Example: Football Kick

• Initial velocity: $25 \frac{m}{s}$ at an angle of $45$ degrees with respect to the horizontal.
• Acceleration: $-9.8 \frac{m}{s^2}$ in the vertical direction.

Time of Flight

• The time of flight depends on the y direction of motion.
• Initial velocity in the y direction: $25 \frac{m}{s} \times \sin(45^\circ) = 17.7 \frac{m}{s}$.
• Final velocity in the y direction: $-17.7 \frac{m}{s}$.
• Using the equation: $v<em>f = v</em>i + at$, solving for time: $t = \frac{v<em>f - v</em>i}{a} = \frac{-17.7 - 17.7}{-9.8} = 3.6 s$.

Horizontal Displacement

• Horizontal velocity is constant.
• Horizontal velocity: $25 \frac{m}{s} \times \cos(45^\circ) = 17.7 \frac{m}{s}$.
• Displacement: $17.7 \frac{m}{s} \times 3.6 s = 64 m$.

Peak Height

• Final velocity at the peak: $0 \frac{m}{s}$.
• Using the equation: $v<em>f^2 = v</em>i^2 + 2a\Delta y$, solving for $\Delta y: \Delta y = \frac{v<em>f^2 - v</em>i^2}{2a} = \frac{0^2 - (17.7)^2}{2 \times (-9.8)} = 16 m$.

Summary

• Constant velocity and constant acceleration concepts can be used to solve projectile motion problems.
• Visualizing the projectile's behavior and using symmetry arguments can simplify problems.
• The acceleration due to gravity is constant near the surface of the Earth.
• The pull of gravity equates to a force, which will be discussed next week.