Projectiles: Launch speed and horizontal distance

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SUMMARY

The discussion focuses on calculating the launch velocity and horizontal range of projectiles in a physics context. The launch velocity for a projectile shot straight upwards with a flight time of 4.70 seconds is determined to be 23.0 m/s using the formula V = -1/2gt, where g is the acceleration due to gravity (-9.8 m/s²). To find the horizontal range when launched at a 45-degree angle, participants suggest using the vertical and horizontal components of the launch speed, incorporating sine and cosine factors for the angle.

PREREQUISITES
  • Understanding of basic kinematics in physics
  • Familiarity with projectile motion concepts
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to apply the equations of motion, specifically d = vit + 1/2gt²
NEXT STEPS
  • Learn how to derive the horizontal range formula for projectile motion
  • Study the effects of launch angle on projectile distance
  • Explore the use of vectors in breaking down launch speed into components
  • Practice problems involving projectile motion with varying angles and speeds
USEFUL FOR

Students in introductory physics courses, educators teaching projectile motion, and anyone seeking to understand the principles of kinematics and projectile trajectories.

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

Homework Statement



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

Homework Equations



d = vit +1/2gt2

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


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

Homework Statement



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

Homework Equations



d = vit +1/2gt2

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

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

Thanks in advance!

Homework Statement


Homework Equations


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 let's you calculate the flight time, the horizontal component with that time let's 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]
 
PeterO said:
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 let's you calculate the flight time, the horizontal component with that time let's you calculate how far away it will land - the range.

How would I do this? I am completely 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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