* Moving in the x and y direction * A projectile is an object shot through the air. This occurs in a parabola curve.

Object dropped Object thrown up Object thrown at an angle projectile- any object that moves through the air or through space, acted on only by gravity (and air resistance, if any)

The vertical acceleration of a projectile is caused by gravity, so a y = -9.8 m/s 2 Parabolic Trajectory

* g remains constant (g= -9.8m/s 2) * a in the x direction is 0 because gravity is not acting on it. * Neglect air resistance * Neglect the effects of the earths rotation

Projectiles launched horizontally

To find how far the ball falls, you use the formula. y =v iy t + 1/2gt 2 1 st second- 5m After 2 seconds- 20m After 3 seconds- 45m The curved path of a projectile produced is a parabola (caused by both horizontal motion and vertical motion. It must accelerate only in the vertical direction)

* The projectile will experience two: * Accelerations (a x = o and a Y = -9.8m/s 2 ) * Velocities * Displacements

Upwardly Launched Projectiles When a projectile is launched at an upward angle, it follows a curved path and finally hits the ground because of gravity. The Vertical distance a cannonball falls below “imaginary path if no gravity” is the same vertical distance it would fall if it were dropped from rest & had been falling for the same amount of time.

* Draw a free body diagram with a coordinate system. * Divide the information into x and y components * Look at your formulas and decided which one(s) to use.

Objects that have been thrown will have a horizontal velocity that stays the same (no horizontal acceleration a x = 0m/s 2) So v fx =v ix in the second formula and third formulas under horizontal motion. (X) Horizontal (Y) Vertical x f- x i = v ix t + ½ a x t 2 y f -y i = v iy t + ½ a y t 2 v fx = v ix + a x t v fy = v iy + a y t v fx 2 - v ix 2 = 2a x (x f- x i ) v fy 2 = v iy 2 + 2a y (y f -y i )

This equation only works when y and y 0 are both the same magnitude y0y0 y

If a ball is thrown up in the air from a moving truck, where will it land? (Ignore air resistance) In front of the truck, behind the truck, or back in the truck

Where will a package land if it is released from a plane? Behind the plane, in front of the plane below the plane

What is the horizontal distance covered by an arrow that was shot through the air at a 60 0 angle with a velocity of 55 m/s? Given v = 55m/s v x =27.5 m/s v yo =47.6m/s a x =0 a y =-9.8m/s 2 t=? x =? Solve V x = cos 60(55m/s)=27.5 m/s V yo = sin60(55m/s)=47.6m/s v y =v yo +a y t x = V x t (we need time) m/s y V x x 0 = 47.6m/s m/s 2 t -47.6m/s = -9.8m/st 4.86 s =t x = 27.5 m/s(9.7s) x = m Total time in the air 4.86s x 2 = 9.7s Need to find time first! To find x dist: x = v x t

* A boat heading due north crosses a river with a speed of 10.0 km/h. The water in the river has a speed of 5.0 km/h due east. Moving frame of reference In general we have (a)Determine the velocity of the boat. (b)If the river is 3.0 km wide how long does it take to cross it?