Projectile Motion (not on Earth)

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Homework Help Overview

The discussion revolves around a projectile motion problem set in a non-Earth environment, specifically involving a truck dropped from a height of 20 meters. Participants are exploring the implications of different gravitational conditions on the motion of the truck.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to determine the acceleration of the truck and its relationship to average speed. Questions arise regarding the use of distance vs. time graphs and the application of kinematic equations in a non-Earth context.

Discussion Status

Some participants have provided insights into the relationship between average speed and acceleration, while others express confusion about the calculations and the application of relevant equations. There is an ongoing exploration of how to approach the problem without reaching a definitive consensus.

Contextual Notes

Participants are working under the assumption that the problem is set in a different gravitational environment, which affects their calculations and understanding of motion. There is mention of a hint provided in the original problem statement regarding the non-Earth context.

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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