# Free-Fall Motion Problem

1. Oct 17, 2007

### pinkcloud

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. The attempt at a solution

2. Oct 17, 2007

### PiratePhysicist

$$a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{(\Delta t)^2}$$
Then do the problem like you would any such problem on Earth! ^_^

3. Oct 17, 2007

### pinkcloud

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?

4. Dec 9, 2011

### kichitakashi

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

5. Dec 9, 2011