Presentation on theme: "Projectile Motion. What is a PROJECTILE? An object that is projected (launched) It continues in motion due to its own inertia, Is only acted upon by gravity."— Presentation transcript:
What is a PROJECTILE? An object that is projected (launched) It continues in motion due to its own inertia, Is only acted upon by gravity No force in the x-direction, only in the y- direction (gravity) Gravity ALWAYS acts in the y-direction, and ONLY the y-direction
So Projectile Motion… Describes the motion of an object in TWO dimensions We will only consider projectiles that stay close to Earth (so a g = -9.8 m/s 2 ) We will continue to disregard air resistance
Some Vocabulary… PROJECTILE – The object being launched/thrown/projected TRAJECTORY – The path that the projectile follows RANGE – The horizontal displacement of the projectile MAXIMUM HEIGHT – The vertical displacement of the projectile at the top of its flight
The Components of Projectile Motion - Velocity We ALWAYS break projectile motion down into its x- and y-components – INITIAL VELOCITY: v i Use cos and sin to find v ix and v iy – v ix = v i cosθ (initial velocity * cos angle) – v iy = v i sinθ (initial velocity * sin angle) vivi v iy v ix
The Components of Projectile Motion - Velocity For a baseball lobbed with an initial velocity of 30 m/s and an angle of 75 degrees, find the initial horizontal and vertical velocities. – Horzontal: x axis – Use cos θ – v ix = v i * cos θ – v ix = 30 * cos 75 – v ix = 30 *.26 – v ix = 7.76 m/s - Vertical: y axis - Use sin θ - v iy = v i * sin θ - v iy = 30 * sin 75 - v iy = 30 *.97 - v iy = 29 m/s
The Components of Projectile Motion - Acceleration ACCELERATION: Always: – a y = -9.8 m/s 2 (vertical acceleration) – a x = 0 m/s 2 (horizontal acceleration) – Why? Think gravity and Newton’s 1 st Law…
Remember Freefall??? Recall that… – a = -9.8 m/s 2, regardless if the object is moving up or moving down – The ONLY force acting on the object is GRAVITY Projectile Motion has the same conditions, and moves in the x-direction simultaneously.
What does this look like? For horizontally projected objects: For objects projected at an angle:
VERY IMPORTANT!!! The components act INDEPENDENTLY of one another!!! When we combine the x- and y-components, we get the characteristic parabola-shape – V x remains constant (a = 0) – V y changes because of gravity (a = -9.8 m/s 2 )
How do I determine the direction? If an object is projected at an angle, the direction is measured from the rightward horizontal
Calculating Projectile Motion - Range Use your kinematic equations! Range is your horizontal displacement - ∆d Remember v x is a constant velocity, so we use:
Projectile Motion – Total Time Once you hit the ground, your motion stops! Time connects your vertical motion with your horizontal motion.
To Calculate Projectile Motion… We use the kinematic equations Remember… Δd = v i t + ½ at 2 – RANGE (x-displacement) uses x-components and total time Δd x = v ix t total + ½ a x t total 2 = v ix t total Remember that a = 0 in the x-direction – HEIGHT (y-displacement) uses y-components Δd y = v iy t + ½ a y t 2 Remember that a = -9.8 m/s 2 in the y-direction
More on Calculations a y = (v fy 2 – v iy 2 ) / 2Δd y – (to find max height) a y = (v fy – v iy ) / t – (to find time to max height) These only apply in the y-direction (v x doesn’t change) a y = -9.8 m/s 2 ALWAYS Remember, at the top of its path, a projectile’s v y = 0 (so v fy = 0)
EXAMPLES A cannonball is fired from a cannon at an angle of 32° and an initial velocity of 54 m/s. A.What are the components of the initial velocity? B.How long does it take the cannonball to reach maximum height? C.What is the cannonball’s maximum height? D.What is the total time of travel? E.What is the cannonball’s range?
A cannonball is fired from a cannon at an angle of 32 ° and an initial velocity of 54 m/s. A. What are the components of the initial velocity?
A cannonball is fired from a cannon at an angle of 32 ° and an initial velocity of 54 m/s. B. How long to maximum height?
A cannonball is fired from a cannon at an angle of 32 ° and an initial velocity of 54 m/s. C. What is the maximum height?
A cannonball is fired from a cannon at an angle of 32 ° and an initial velocity of 54 m/s. D. What is the total time?
A cannonball is fired from a cannon at an angle of 32 ° and an initial velocity of 54 m/s. E. What is the range?