Projectiles
When an object is thrown into the air and is subject to gravitational acceleration, it is termed a projectile. The fascinating trajectory of a projectile can be dissected into horizontal and vertical motions, each of which can be independently scrutinised using the equations of motion.
Horizontal Motion of Projectiles
In the horizontal motion of projectiles, it's pivotal to note that the velocity remains constant. This is because gravitational acceleration does not influence it, and air resistance is often neglected in basic physics.
- Uniform Velocity: The horizontal velocity does not change throughout the motion.
- Equation of Horizontal Motion: The horizontal distance (s) is calculated by multiplying the initial horizontal velocity (u) with the time of flight (t), expressed as s = ut.
Vertical Motion of Projectiles
Contrastingly, the vertical motion of projectiles is subject to gravitational acceleration, resulting in a non-uniform velocity.
- Accelerated Motion: The vertical velocity changes due to gravitational acceleration.
- Equations of Vertical Motion:
- Final velocity (v) is obtained by adding the product of gravitational acceleration (a) and time (t) to the initial velocity (u): v = u + at.
- Vertical distance (s) is calculated by multiplying initial velocity (u) with time (t) and adding half of the product of gravitational acceleration (a) and square of time (t2): s = ut + 0.5at2.
- The square of final velocity (v2) is obtained by adding twice the product of gravitational acceleration (a) and vertical distance (s) to the square of initial velocity (u2): v2 = u2 + 2as.
Maximum Height and Range
- Maximum Height (H): The highest point reached by a projectile, calculated by setting the final vertical velocity (v) to zero and rearranging the third equation of vertical motion.
- Range (R): The total horizontal distance covered by a projectile, calculated using the horizontal motion equation.
Example Question 1
A ball is kicked with an initial velocity of 20 m/s at an angle of 45 degrees to the horizontal. Calculate the maximum height reached by the ball.
Solution:
- Initial vertical velocity (u) = 20sin45 = 14.14 m/s
- Gravitational acceleration (a) = -9.8 m/s2 (negative as it acts downward) Using the equation v2 = u2 + 2as, where v = 0 at the maximum height, we find the maximum height (s).
Example Question 2
Calculate the range of the projectile in the above example.
Solution:
- Initial horizontal velocity (u) = 20cos45 = 14.14 m/s
- Time of flight (t) can be found using vertical motion equations and then used in the horizontal motion equation s = ut to find the range.
Relative Motion
Relative motion is the concept of analysing the motion of an object from the perspective of another moving object, which is termed the frame of reference. The relative velocity is the velocity of one object as observed from another moving object.
Basic Concepts
- Relative Velocity: Defined as VAB = VA - VB, where VAB is the velocity of object A relative to object B, VA is the velocity of A, and VB is the velocity of B.
- Same Direction Motion: If two objects move in the same direction, their relative velocity is the difference of their velocities.
- Opposite Direction Motion: If two objects move in opposite directions, their relative velocity is the sum of their velocities.
Example Question 3
Two cars are moving in the same direction along a straight road. Car A is travelling at 20 m/s and Car B at 15 m/s. What is the velocity of Car A relative to Car B?
Solution: Since they are moving in the same direction, VAB = VA - VB = 20 m/s - 15 m/s = 5 m/s.
Example Question 4
If Car A is moving at 20 m/s to the east and Car B is moving at 15 m/s to the west, what is the velocity of Car B relative to Car A?
Solution: Since they are moving in opposite directions, VBA = VB + VA = 15 m/s + 20 m/s = 35 m/s to the west.
In these notes, we've delved into the applications of kinematic equations in projectiles and relative motion, providing a foundational understanding of these concepts with practical examples. The example questions are crafted to mirror potential exam questions, ensuring students are well-equipped to apply theoretical knowledge in an exam setting.
FAQ
The launch speed of a projectile significantly impacts its subsequent motion. A greater launch speed increases both the maximum height and range of the projectile, assuming the launch angle remains constant. The horizontal and vertical components of the velocity, obtained by resolving the initial velocity vector, determine the time of flight and peak height of the projectile. A higher launch speed means a greater initial kinetic energy, allowing the projectile to ascend higher and travel further before gravitational potential energy and, if considered, air resistance bring it to a stop.
In riverboat problems, relative motion is crucial to determine the resultant path (or resultant velocity) of the boat. The velocity of the boat relative to the water and the velocity of the water (river) itself combine to give the actual velocity of the boat observed from the shore. Vector addition is used to find this resultant velocity. If a boat aims straight across a flowing river, it will drift downstream due to the river’s current. To counteract this, the boat must steer at an upstream angle, with the exact angle determined by the relative speeds of the boat and the river, ensuring it reaches the opposite bank directly across from its starting point.
The classical equations of motion are derived for inertial (non-accelerating) frames of reference. In non-inertial (accelerating) frames, additional "fictitious" forces, such as the centrifugal force or Coriolis force, need to be considered to accurately describe motion. These fictitious forces arise due to the acceleration of the frame of reference and do not have a physical origin in the way gravitational or electromagnetic forces do. To analyse motion in non-inertial frames, either the fictitious forces must be included in the Newtonian equations of motion, or the equations of motion must be modified to account for the accelerating frame of reference.
The angle of projection plays a crucial role in determining the range and height of a projectile. For a given initial velocity, a projectile launched at an angle of 45 degrees will achieve the maximum range, assuming no air resistance. As the launch angle increases from 0 to 45 degrees, both the range and maximum height increase. Beyond 45 degrees, increasing the angle further will continue to increase the maximum height but decrease the range. This is because the vertical component of the velocity increases, allowing the projectile to reach higher altitudes, while the horizontal component decreases, reducing the horizontal distance covered.
Air resistance, often neglected in basic physics problems, can significantly alter the trajectory of a projectile. When a projectile moves through the air, it experiences a resistive force proportional to its velocity. This force acts opposite to the direction of motion, causing the projectile to decelerate. In the vertical motion, this means the projectile will reach its maximum height sooner and will not ascend as high as it would in a vacuum. Horizontally, the projectile will not travel as far as expected due to the additional resistive force. The exact impact of air resistance is determined by various factors, including the shape, size, and speed of the projectile, and requires more advanced mathematical models to accurately predict.
Practice Questions
To find the maximum height attained by the projectile, we first need to determine the vertical component of the initial velocity, which can be found using trigonometry. The initial vertical velocity (u) is given by u = 20sin(30) = 10 m/s. We know that at the maximum height, the final vertical velocity is 0 m/s. Using the equation v2 = u2 + 2as, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s2, since it is acting downward), and s is the maximum height. Rearranging for s gives us s = (v2 - u2) / (2a). Substituting in the known values, we find s = (0 - 100) / (-19.6) = 5.1 m. Thus, the projectile reaches a maximum height of 5.1 m.
To find the time before the trains collide, we need to determine their relative velocity. Since they are moving towards each other, we add their velocities together: Vrel = VA + VB = 40 m/s + 30 m/s = 70 m/s. The time before collision (t) can be found by dividing the initial distance between them (s) by their relative velocity: t = s / Vrel = 2100 m / 70 m/s = 30 s. Therefore, the trains have 30 seconds before they collide. This problem demonstrates the application of relative motion concepts in a practical scenario, requiring the understanding of relative velocity and the ability to apply it in a real-world context.
