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## Projectile Motion

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### Projectile Motion

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:

 xV[sub]x[/sub] = ?Δx = ?Δt = ? ya[sub]g[/sub] = -9.80 m/s2V[sub]1y[/sub] = 0Δy = ?Δt = ?V[sub]2y[/sub] = ?

Δt for both x and y are the same

2. Use $V_x=\frac{\Delta x}{\Delta t}$ with the x column and/or the kinematic equations w/ the y column

Launching and Landing at Same Height Problem Solving Strategy

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

2.
 xV[sub]x[/sub] = ?Δx = ?Δt = ? ya[sub]g[/sub] = -9.80 m/s2V[sub]1y[/sub] = ?Δy = 0 *if position 2 at same height as launchV[sub]2y[/sub] = 0 *if position 2 at highest point

3. Analyze x direction using $V_x=\frac{\Delta x}{\Delta t}$ and/or the y direction using the kinematic equations.
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Jessica
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Sapphire

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### Re: Projectile Motion

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]

$x(t) = (v_0cos\theta_0)t$
$y(t) = (v_0sin\theta_0)t-\frac{1}{2}gt^2$
• 0

Tales of Ostlea :
Dragon Cave :
Magistream :

Jessica
Board Owner
Topic Author

Sapphire

Posts: 3,469
Topics: 1,244
Articles: 30
Joined: December 22nd, 2010, 8:04 pm