# Homework Help: Linear Motion

1. Nov 26, 2006

### ganon00

Neglecting air resistance, calculate the position (in meters) and velocity
(in m/s) of a golf ball that is dropped from the empire state building after 1.0s, 2.0s, and 3.0 s with the release point.:surprised

2. Nov 26, 2006

### Hootenanny

Staff Emeritus
One is expected to some some effort in answering the question...

3. Nov 26, 2006

### ganon00

all i need is the equation to solve this problem...... not the answer...

i'm partially leaning toward, v=delta(d)/delta(t) a=delta(v)/delta(t)

4. Nov 26, 2006

### Hootenanny

Staff Emeritus
After the ball is release, what force(s) are acting on the ball?

5. Nov 26, 2006

### ganon00

9.8m/s, which is gravity

Last edited: Nov 26, 2006
6. Nov 26, 2006

### Hootenanny

Staff Emeritus
That is an acceleration, not a force, but it was where I was leading to so no worries. Now, if an object starts at rest and accelerates at 9.8m/s2 (note the squared as opposed to your units) how fast will be be travelling after one second?

7. Nov 26, 2006

### ganon00

9.8meters after 1 second
then is it doubled and tripled after 2 and 3 seconds

8. Nov 26, 2006

### Hootenanny

Staff Emeritus
I said how fast not how far.

9. Nov 26, 2006

### ganon00

how fast.....9.8m/s

this is where im stuck i dont know....... terminal velocity im guessing

so possibly 75mi/hr after 5 seconds

Last edited: Nov 26, 2006
10. Nov 26, 2006

### Hootenanny

Staff Emeritus
Yes, that is correct. Have you met kinematic equations before? If not are you comfortable with energy conservation?

11. Nov 26, 2006

### ganon00

no i do not know kinematic equations, but i do know a little about energy conservation...

note: i have only been familiar with physics for about a week now.

0.5m/s after 1.0seconds.... possibly

12. Nov 26, 2006

### ganon00

i got 0.5m/s by solving for acceleration by dividing the change in velocity/over the time taken

13. Nov 26, 2006

### ganon00

i crunched a few numbers and im coming up with this for the answer is this correct:

after 1.0seconds its speed is 0.5m/s
after 2.0seconds its speed is 1.0m/s
after 3.0seconds its speed is 1.5m/s

14. Nov 26, 2006

### Hootenanny

Staff Emeritus
What are you doing? You have given me the correct answer in one of your above posts;

after 1.0 second speed = 9.8m/s
after 2.0 seconds speed = 2 x 9.8 m/s
after 3.0 seconds speed = 3 x 9.8 m/s

Do you follow?

15. Nov 26, 2006

### ganon00

so all i have to do to get the answer is to multiply 9.8 by two and 3

there has to be more than that

im really trying to understand

16. Nov 26, 2006

### Ja4Coltrane

there is not more than that--you are over complicating the simple concept of acceleration. 1m/s^2 means one meter per second PER SECOND. which means that every second it is going one meter per second faster. so 9.8 m/(ss) means 9.8 after one second, 9.8*2 after two and so on.
do you know any calculus?

17. Nov 26, 2006

### crimsondarkn

Yeah...this is an incredibly easy physics question , you're stressing too much over it.

18. Nov 26, 2006

### turdferguson

acceleration = velocity/time
velocity = acceleration * time
Thats all there is to the velocity part. Because you are neglecting air resistance, the ball will continue to accelerate infinitely with no terminal velocity.
Distance is a little trickier, you need the kinematic d=vo+.5at^2, which simplifies to 1/2at^2 in your case. When you do calculus, youll realize that this relationship between distance and acceleration is an anti-derivitive and theres a reason for that. But for now, just plug and chug

19. Nov 26, 2006

### ganon00

ok i think then i got the answer to you the prob...

after 1.0seconds its speed is 9.8m/s
after 2.0seconds its speed is 19.6m/s
after 3.0seconds its speed is 29.4m/s

and i haven't taken calculus yet just algebra 2 I'm a freshman in high school.