Position & Velocity of Golf Ball Dropped from Empire State Building

[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.

 

[Hootenanny]

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

 

[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)

 

[Hootenanny]

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

 

[ganon00]

9.8m/s, which is gravity

 

[Hootenanny]

im guessing 9.8m/s, which is gravity

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 traveling after one second?

 

[ganon00]

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

 

[Hootenanny]

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

I said how fast not how far.

 

[ganon00]

how fast...9.8m/s

this is where I am stuck i don't know... terminal velocity I am guessing

so possibly 75mi/hr after 5 seconds

 

[Hootenanny]

how fast...9.8m/s

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

 

[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

 

[ganon00]

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

 

[ganon00]

i crunched a few numbers and I am 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

 

[Hootenanny]

i crunched a few numbers and I am 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

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?

 

[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 thatim really trying to understand

 

[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?

 

[crimsondarkn]

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

 

[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-derivative and there's a reason for that. But for now, just plug and chug

 

[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.