1. Motion of Projectiles

2. An object that has projectile motion has an initial velocity, follows a curved path(trajectory) and reaches the ground due to the force of gravity. Since the object follows a curved path, its motion will have both a horizontal and a vertical component.

3. The vertical and horizontal motions are separate from one another. In the horizontal direction there is no acceleration, and therefore the motion is described as URM; this means that the velocity along the horizontal remains unchanged. In the vertical direction the object’s motion is influenced by gravity and therefore has an acceleration equal to -9.8 m/s 2.

4. The variables that need to be considered in projectile motion are: time (t), the horizontal position (x), the vertical position, (y), the horizontal velocity (v x ), the vertical velocity (v y ) and the vertical acceleration (a y ).

5. Equations for Projectile Motion A.Motion along the horizontal: x f = x i + v ix Δt B.Motion along the vertical: 1.v fy = v iy + a y Δt 2.y f = y i + ½ (v iy + v fy )Δt 3.y f = y i + v iy Δt + ½ a y (Δt) 2 4.v fy 2 = v iy 2 + 2a y Δy **DO NOT MIX VELOCITIES. v i in the horizontal direction can not be substituted in for v i in the vertical direction. The only variable that can be used in both directions is t, time.

6. Launch Direction A. When an object is launched horizontally, the initial velocity along the vertical is 0 m/s, v iy = 0 m/s Example: A ball is thrown horizontally from a height of 200.0 m with a velocity of 3.0 m/s. What is the ball’s position at t= 2.0 s?

7. t i = 0 s x i = 0m y i = 200m v ix = 3.0 m/s v iy =0m/s t f = 2 s x f = ? y f = ? v fx = 3.0m/s v fy = ? a x = 0 m/s 2 a y = -9.8 m/s 2 Motion along the horizontal: x f = x i + v ix Δt = 0 m + 3.0 m/s x 2 s = 6 m Motion along the vertical: y f = y i + v iy Δt + ½ a y (Δt) 2 = 200 m + 0x2s + ½ (-9.8 m/s 2 )(2 s) 2 = 180.4 m

9. Projectiles Launched at an angle B.When an object is launched at an angle we must use trigonometric functions to determine both v iy and v ix. v ix = v i x cos θ i v iy = v i x sin θ i vivi v iy v ix θiθi

10. Range The range of an object launched at an angle is equivalent to its total horizontal distance. It can be found using: Range = v i 2 sin2 θ i g **The range can only be determined if the initial and final vertical heights remain unchanged.