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Trajectory: the path of an object undergoing projectory motion

3 types:

• horizontal launch
• launching and landing at same height
• launching and landing at different heights

Horizontal Launch Problem Solving Strategy

1. Separate the x and y variables:

x V[sub]x[/sub] = ? Δx = ? Δt = ?

y a[sub]g[/sub] = -9.80 m/s2 V[sub]1y[/sub] = 0 Δy = ? Δt = ? V[sub]2y[/sub] = ?

Δt for both x and y are the same


2. Use with the x column and/or the kinematic equations w/ the y column

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Launching and Landing at Same Height Problem Solving Strategy

1. Calculate the x and y components of the initial velocity.

2.

x V[sub]x[/sub] = ? Δx = ? Δt = ?

y a[sub]g[/sub] = -9.80 m/s2 V[sub]1y[/sub] = ? Δy = 0 *if position 2 at same height as launch V[sub]2y[/sub] = 0 *if position 2 at highest point

3. Analyze x direction using and/or the y direction using the kinematic equations.

If air resistance is neglected, the motion of a projectile is the superposition of two independent motions:

• a horizontal motion with constant velocity (equal to initial horizontal velocity, v[sub]0x[/sub] = v[sub]0[/sub]cosθ[sub]0[/sub];
• a vertical motion with constant acceleration (−g) and initial vertical velocity v[sub]0y[/sub] = v[sub]0[/sub]sinθ[sub]0[/sub]