Projectile Motion

Definition and Concept:

Projectile motion is the motion of an object in two dimensions, under the influence of gravity. The object is projected into the air, and moves along a curved path, known as a parabola. The motion of the object can be analyzed using the principles of kinematics and the laws of motion.

Variables:

The motion of a projectile can be described in terms of several variables:

• Initial velocity ($v_0$): The velocity with which the projectile is launched.
• Angle of projection ($θ$): The angle at which the projectile is launched.
• Maximum height ($H$): The maximum height reached by the projectile.
• Time of flight ($t$): The total time for which the projectile is in the air.
• Velocity components: The vertical and horizontal components of the projectile’s velocity.

Equations:

The motion of a projectile can be described using a set of equations. Some of the key equations are:

• Vertical displacement: $y = v_0\sin(θ)*t - (1/2)gt^2$
• Horizontal displacement: $x = v_0\cos(θ)*t$
• Maximum height: $H = (v_0\sin(θ))^2/(2*g)$
• Time of flight: $t = 2*v_0\cos(θ)/g$

Tips for Problem Solving:

To solve problems related to projectile motion, you should follow these steps:

• Identify the variables that are given in the problem and those that need to be calculated.
• Use the relevant equations to solve for the required variables.
• Pay attention to the signs of the variables, especially when dealing with the vertical displacement and velocity components.
• Make sure that the units are consistent throughout the calculations.
• Check your final answer for correctness and units.

Examples

Example 1: A ball is thrown from a height of 5 meters above the ground with an initial speed of 10 m/s at an angle of 30 degrees above the horizontal. Find the time of flight, and the maximum height.

Solution:

• The vertical velocity component is v0y = v0_sin(θ) = 10_sin(30) = 5 m/s
• The horizontal velocity component is v0x = v0_cos(θ) = 10_cos(30) = 8.7 m/s
• The time of flight is t = 2_v0y/g = 2_5/9.8 = 1.02 s
• The maximum height is H = (v0y)^2/(2_g) = (5^2)/(2_9.8) = 1.27 m

Example 2: A cannonball is fired from a cannon at an initial velocity of 100 m/s at an angle of 45 degrees above the horizontal. How far does the cannonball travel before hitting the ground?

Solution:

• The horizontal velocity component is v0x = v0_cos(θ) = 100_cos(45) = 70.7 m/s
• The time of flight is t = 2_v0y/g = 2_(100*sin(45))/9.8 = 14.3 s
• The range is R = v0x_t = 70.7_14.3 = 1012 m