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Projectiles: Launch speed and horizontal distance

 
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Sep14-11, 07:27 PM   #1
 

Projectiles: Launch speed and horizontal distance


New to the forums, just starting a physics 11 course online and having some troubles with formulas.

1. The problem statement, all variables and given/known data

Projectile shot straight upwards, flight time = 4.70 sec
Second Launch at 45 degrees (horizontal range approx. 50-60 meters

A) Find Launch Velocity
B) Find Horizontal Range when shot at 45 degrees

2. Relevant equations

d = vit +1/2gt2

V = - 1/2gt ("made" this one myself from above formula, unsure if correct)


3. The attempt at a solution


Launch velocity


V = - 1/2gt
g = -9.8m/s
t = 4.7s

V = -[((1/2(-9.8))(4.7)]
V = -((-4.9)(4.7))
V = 23.0 m/s


I have no idea how to calculate the horizontal range, I need a formula that incorperates the 45 degrees

Thanks in advance!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
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Sep14-11, 07:38 PM   #2
 
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Quote by gratsoy View Post
New to the forums, just starting a physics 11 course online and having some troubles with formulas.

1. The problem statement, all variables and given/known data

Projectile shot straight upwards, flight time = 4.70 sec
Second Launch at 45 degrees (horizontal range approx. 50-60 meters

A) Find Launch Velocity
B) Find Horizontal Range when shot at 45 degrees

2. Relevant equations

d = vit +1/2gt2

V = - 1/2gt ("made" this one myself from above formula, unsure if correct)


3. The attempt at a solution


Launch velocity


V = - 1/2gt
g = -9.8m/s
t = 4.7s

V = -[((1/2(-9.8))(4.7)]
V = -((-4.9)(4.7))
V = 23.0 m/s


I have no idea how to calculate the horizontal range, I need a formula that incorperates the 45 degrees

Thanks in advance!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
You can work from first principles: Calculate the vertical and horizontal components of the launch speed - using sin45 and cos 45 factors.
The vertical component lets you calculate the flight time, the horizontal component with that time lets you calculate how far away it will land - the range.
Many Physics texts will have a section where a formula is derived, so you can simply substitute the 23 m/s to get an answer [assuming 23 m/s is correct - it certainly is a reasonable answer]
 
Sep15-11, 06:46 PM   #3
 
Quote by PeterO View Post
You can work from first principles: Calculate the vertical and horizontal components of the launch speed - using sin45 and cos 45 factors.
The vertical component lets you calculate the flight time, the horizontal component with that time lets you calculate how far away it will land - the range.
How would I do this? I am completly new to this and need a formula with a simple step-by-step guide on how to implement that formula to derive the answer.
 
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angle, horizontal distance, launch velocity, velocity
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