Ideal Projectile Motion Section 10.4a

Ideal Projectile Motion What do we mean by the word “ideal” here? Assumptions: The projectile behaves like a particle moving in a vertical coordinate plane. The only force acting on the projectile during its flight is the constant force of gravity. The projectile will follow a trajectory that is perfectly parabolic.

Ideal Projectile Motion Position, velocity, acceleration due to gravity, and launch angle at t = 0… Initial velocity: Initial position: r = 0 at time t = 0

Ideal Projectile Motion Position, velocity, and acceleration at a later time t… Horizontal range

Ideal Projectile Motion Newton’s second law of motion: the force acting on the projectile is equal to the projectile’s mass times its acceleration. For ideal motion, this force is solely the gravitational force: Let’s solve this initial value problem, using initial conditions: and when First integration: Second integration:

Ideal Projectile Motion Previous equations: Substitution:

Ideal Projectile Motion This is the vector equation for ideal projectile motion. The angle is the projectile’s launch angle (firing angle, angle of elevation), and is the projectile’s initial speed. Break into a pair of scalar equations: These are the parametric equations for ideal projectile motion. Gravity constants:

Ideal Projectile Motion If the ideal projectile is fired from the point instead of the origin:

Practice Problem Ten seconds after firing, the projectile is about A projectile is fired from the origin over horizontal ground at an initial speed of 500 m/sec and a launch angle of 60 degrees. Where will the projectile be 10 sec later? Parametric equations for this situation, evaluated at this time: Ten seconds after firing, the projectile is about 3840.127 m in the air and 2500 m downrange

Height, Flight Time, and Range The projectile reaches its highest point when its vertical velocity component is zero: Height at this time:

Height, Flight Time, and Range To find the total flight time of the projectile, find the time it takes for the vertical position to equal zero:

Height, Flight Time, and Range To find the projectile’s range R, the distance from the origin to the point of impact on horizontal ground, find the value of x when t equals the flight time: Note: The range is largest when or .

Height, Flight Time, and Range For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed and launch angle : Maximum Height: Flight Time: Range:

Practice Problems m sec m Find the maximum height, flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m/sec and a launch angle of 60 degrees. Then graph the path of the projectile. Max Height: m Flight Time: sec Range: m

Practice Problems Graph window: [0, 25000] by [–2000, 10000] Find the maximum height, flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m/sec and a launch angle of 60 degrees. Then graph the path of the projectile. Parametric equations for the path of the projectile: Graph window: [0, 25000] by [–2000, 10000]

Practice Problems To open the 1992 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow. Suppose that Rebollo shot the arrow at a height of 6 ft above ground level 30 yd from the 70-ft-high cauldron, and he wanted the arrow to reach a maximum height exactly 4 ft above the center of the cauldron.

Practice Problems Express in terms of the initial speed and firing angle . (b) Use ft and the result from part (a) to find the value of .

Practice Problems (c) Find the value of . Values when the arrow reaches maximum height:

Practice Problems (d) Find the initial firing angle of the arrow. Put these parts together: