# Free-Fall Motion Problem

## 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?

## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
$$a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{(\Delta t)^2}$$
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

PeterO
Homework Helper

## 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?

## The Attempt at a Solution

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: