Welcome

It is highly recommended that you have Javascript enabled; many features will not work unless you do. The recommended browsers are Firefox and Chrome; the board is also NOT mobile-friendly.

If you were referred by someone (ex. me, Jessica) please put their username in the referral box on the registration page. Ask them if you don't know their username.

If you are visiting for TESTING PURPOSES ONLY, this is the test account information:

Test

Select a forum to post in:

## Projectile Motion

Science topics

Moderators: Jessica, Teachers

Looking for a topic?
View the Science Topics Directory to find topics quicker under different kinds of science.

### 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.
• 0

Tales of Ostlea :
Dragon Cave :
Magistream :

Jessica
Board Owner
Topic Author

Sapphire

Posts: 3,468
Topics: 1,244
Articles: 30
Joined: December 22nd, 2010, 8:04 pm

### 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: 3,468
Topics: 1,244
Articles: 30
Joined: December 22nd, 2010, 8:04 pm