How Do You Calculate Gravity and Final Velocity on Jupiter?

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Homework Help Overview

The problem involves calculating the acceleration due to gravity and the final velocity of a steel ball dropped from a height of 300 meters on Jupiter, taking 5 seconds to reach the ground. The context is set in a futuristic scenario on Jupiter.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between distance, time, and acceleration, with some attempting to apply equations of motion. Questions arise regarding how to determine change in velocity and time, as well as the interpretation of given values.

Discussion Status

Some participants have provided guidance on using kinematic equations to approach the problem, while others express confusion about specific components of the calculations. Multiple interpretations of the problem are being explored, particularly regarding the average and final velocities.

Contextual Notes

Participants note the absence of initial attempts at solutions from some users, and there is a recurring emphasis on understanding the underlying concepts rather than simply calculating values.

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Homework Statement



A research party on Jupiter in the year 3005 drops a steel ball from 300 meters above ground. It takes exactly 5 seconds to reach the ground.

a. Calculate the value of g (the acceleration due to gravity) on the surface of Jupiter.
b. What is the velocity of the steel ball the instant before it strikes the ground?

Homework Equations


.dont' know how to do it...please help


The Attempt at a Solution



None yet...dont' know how to do it...please help
 
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Start with
a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{(\Delta t)^2}
Then do the problem like you would any such problem on Earth! ^_^
 
Thanks...but I'm confused about the part that it give the distance=300 meters; 5 sec=the time. how do i get the change in velocity and the change in time?
 
given:
h = 300m
t = 5s
Vi = 0 m/s
a=?

use the following equation: h= (Vi)(t) + (at2)/2
substitute the given values: 300m = (0m/s)(5s) + [a(5s)2]/2
then we get: 300m = 25s2a/2
a = 24m/s2
 
pinkcloud said:

Homework Statement



A research party on Jupiter in the year 3005 drops a steel ball from 300 meters above ground. It takes exactly 5 seconds to reach the ground.

a. Calculate the value of g (the acceleration due to gravity) on the surface of Jupiter.
b. What is the velocity of the steel ball the instant before it strikes the ground?

Homework Equations


.dont' know how to do it...please help

The Attempt at a Solution



None yet...dont' know how to do it...please help

Since you have already been given the answer - here is the conceptual way to answer.

Basis: When a body undergoes constant acceleration from zero, the average velocity is one half of the final velocity.

Here: 300m covered in 5 seconds → average velocity is 60 m/s

Thus final velocity - just before hitting Jupiter is 120 m/s [part (b) done]

In order to reach that velocity in 5 seconds, you must add 24 m/s each second, so the acceleration due to gravoity on Jupiter would appear to be 24 m/s2/. [part (a)]
 
Last edited:

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