Projectile Motion (not on Earth)

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SUMMARY

The discussion focuses on solving a projectile motion problem in a non-Earth environment, specifically involving a truck dropped from a height of 20 meters. The relevant equation used is delta y = vt + 1/2at², where the user attempts to calculate the truck's height after 3 seconds. The user initially calculates an acceleration of 1 m/s² but later finds a discrepancy using the same equation, resulting in a height of 15.86 meters after 3 seconds. The conversation emphasizes the importance of using SUVAT equations for accurate calculations in physics.

PREREQUISITES
  • Understanding of basic kinematics and projectile motion
  • Familiarity with SUVAT equations
  • Knowledge of distance vs. time graphs
  • Ability to perform calculations involving acceleration and average speed
NEXT STEPS
  • Study the derivation and application of SUVAT equations in different gravitational contexts
  • Learn how to create and interpret distance vs. time graphs for projectile motion
  • Explore the effects of varying acceleration on projectile motion
  • Investigate the implications of non-Earth gravitational environments on motion calculations
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Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion in non-standard gravitational conditions.

randomphysicsguy123
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Homework Statement
A truck is dropped out of a building that is 20 meters high. When the truck started to fall, it went 1 meter after the first second. After a total of 3 seconds, how high is the truck located from the ground? (Hint: This is not on earth)
Relevant Equations
delta y = vt+1/2at^2
I know I need to solve for acceleration as I am not on Earth and I am assuming I should create a distance vs. time graph. But overall I am unsure what to do. Please help been stuck on this for a while.
 
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randomphysicsguy123 said:
Homework Statement:: A truck is dropped out of a building that is 20 meters high. When the truck started to fall, it went 1 meter after the first second. After a total of 3 seconds, how high is the truck located from the ground? (Hint: This is not on earth)
Relevant Equations:: delta y = vt+1/2at^2

I know I need to solve for acceleration as I am not on Earth and I am assuming I should create a distance vs. time graph. But overall I am unsure what to do. Please help been stuck on this for a while.
One metre in one second is an average speed of ##1m/s##, right?
 
Correct. But could you provide a bit more insight as what I should do I am really lost.
 
randomphysicsguy123 said:
Correct. But could you provide a bit more insight as what I should do I am really lost.
How is average speed related to acceleration?
 
average speed=d/t
a=change in v/ change in time
 
randomphysicsguy123 said:
average speed=d/t
a=change in v/ change in time
Can you use those to find the acceleration on the mystery planet?
 
Yes I could however I keep getting 1m/s^2 but when xf=xi+vit+1/2a*t2 to check if that statement is true I get .92m/s^2
 
randomphysicsguy123 said:
Yes I could however I keep getting 1m/s^2 but when xf=xi+vit+1/2a*t2 to check if that statement is true I get .92m/s^2
If the acceleration is ##1m/s^2##, how far does the object fall in the first second?
 
1 meter
 
  • #10
randomphysicsguy123 said:
1 meter
If that was a guess it was wrong. Why not use ##s = ut + \frac 1 2 a t^2##, with ##u = 0## and ##t = 1s##?

Why guess when you have SUVAT equations?
 
  • #11
Didn't think about that. Regardless, I already turned it in and I used the equation xf=xi+vit+1/2a*t^2, which got me 15.86 meters at 3 seconds. Hope it was right. Thank you for your help.
 
  • #12
randomphysicsguy123 said:
Didn't think about that. Regardless, I already turned it in and I used the equation xf=xi+vit+1/2a*t^2, which got me 15.86 meters at 3 seconds. Hope it was right. Thank you for your help.
xf=xi+vit+1/2a*t^2 tells you that, starting from rest, the distance is proportional to t2. So after three times the time, how many times the distance?
 
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