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**Ideal Projectile Motion**

Section 10.4a

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**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.

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**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

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**Ideal Projectile Motion**

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

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**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:

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**Ideal Projectile Motion**

Previous equations: Substitution:

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**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:

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**Ideal Projectile Motion**

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

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**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 m in the air and 2500 m downrange

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**Height, Flight Time, and Range**

The projectile reaches its highest point when its vertical velocity component is zero: Height at this time:

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**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:

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**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

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**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:

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**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

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**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]

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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.

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**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

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**Practice Problems (c) Find the value of .**

Values when the arrow reaches maximum height:

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**Practice Problems (d) Find the initial firing angle of the arrow.**

Put these parts together:

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Physics Lesson 6 Projectile Motion

Physics Lesson 6 Projectile Motion

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