Projectile Motion Introduction Horizontal launch.

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Presentation transcript:

Projectile Motion Introduction Horizontal launch

#1 A ball is dropped from a height of 125 m. How long does it take the ball to strike the ground? What is the velocity of the ball at the ground? #2 A ball rolls at 15 m/s for 5 seconds. How far does the ball roll? INCLUDE A LIST OF GIVENS FOR EACH motion map

Projectile motion Projectile – an object with only gravity acting on it No air resistance (in our examples) No propellant After some initial condition (throwing, etc)

Projectile motion vo Which ball hits the ground first? A: dropped B: launched horizontally vo A B

Projectile motion Perpendicular vectors are independent They do not directly affect each other Break all 2-D motion into two separate vector components. Choose easy components that make sense

Projectile motion Gravity pulls down (vertical vector), so it has no effect on the horizontal vector ***This (vertical) should be a dimension that we analyze “Horizontal” would then be the other (┴)

Projectile motion Acceleration in the vertical (y) direction will continue to be g = -10 m/s2 ay = -10 m/s2 If air resistance is ignored, there is nothing changing the horizontal velocity* ax = 0 m/s2 vox = vx *This may change in future problems

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: Vertical (y) Constant acceleration (ay = -10 m/s2) Equation:

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: dx = d0x + vx·t Vertical (y) Constant acceleration (ay = -10 m/s2) Equation:

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: dx = d0x + vx·t Vertical (y) Constant acceleration (ay = -10 m/s2) Equation: dy = … vy = …

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: dx = d0x + vx·t Determines: ____ Vertical (y) Constant acceleration (ay = -10 m/s2) Equation: dy = … vy = … Determines: ___

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: dx = d0x + vx·t Determines: ____ Vertical (y) Constant acceleration (ay = -10 m/s2) Equation: dy = … vy = … Determines time

Constant acceleration Projectile motion Horizontal (x) Constant velocity (ax = 0 m/s2) vox = vx Equation: dx = d0x + vx·t Determines range Vertical (y) Constant acceleration (ay = -10 m/s2) Equation: dy = … vy = … Determines time

A ball is rolled horizontally off a 125 meter ledge at a speed of 15 m/s. What is the range of the ball? (How far will it go horizontally, dx) What is the velocity when it hits the ground? magnitude and direction (angle)

Projectile Motion Horizontal launch – v0y = 0 m/s

Projectile Motion Angled launch

#1 A ball is thrown upward at 30 m/s from an 80 meter cliff. How long does it take the ball to strike the ground? What is the velocity of the ball at the ground? #2 A ball rolls at 40 m/s for 8 seconds. How far does the ball roll?

Projectile motion v0 v0 vx v0y When starting with an angled velocity, show the velocity as a sum of perpendicular components horizontal (x), and vertical (y) v0 v0 vx v0y

Projectile motion v0 v0y v0y vx vx Remember, vectors can be moved around v0 vx v0y vx v0y

A ball is launched off an 80 meter cliff. The initial velocity of the ball is 50 m/s at an angle of 37º above horizontal. What is the range of the ball? What is the velocity when it hits the ground?

Motion Map

Motion Map

Motion Map

Projectile Motion Horizontal launch – v0y = 0 m/s Angled launch Always find vx and voy first Never use the total/vector sum in equations Remember: vy = 0 m/s at max vx is constant position has two coordinates (dx, dy)