# Position & Velocity of Golf Ball Dropped from Empire State Building

• ganon00
In summary, a golf ball is dropped from the Empire State Building after 1.0s, 2.0s, and 3.0s. After 1.0s, the ball's speed is 0.5m/s. After 2.0s, the ball's speed is 1.0m/s. After 3.0s, the ball's speed is 1.5m/s.
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.

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

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)

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

9.8m/s, which is gravity

Last edited:
ganon00 said:
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?

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

ganon00 said:
9.8meters after 1 second
then is it doubled and tripled after 2 and 3 seconds
I said how fast not how far.

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

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ganon00 said:
how fast...9.8m/s
Yes, that is correct. Have you met kinematic equations before? If not are you comfortable with energy conservation?

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

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

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

ganon00 said:
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?

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

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?

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

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

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.

## 1. What is the initial position and velocity of the golf ball when dropped from the Empire State Building?

The initial position of the golf ball is at the top of the Empire State Building, which is approximately 1,250 feet above the ground. The velocity of the golf ball when dropped is 0 feet per second, since it is initially at rest.

## 2. How long will it take for the golf ball to reach the ground?

The time it takes for the golf ball to reach the ground can be determined using the equation t = √(2h/g), where t is time, h is the initial height, and g is the acceleration due to gravity (9.8 m/s²). Plugging in the values for the Empire State Building (381 meters) and solving for t, we get a time of approximately 8.8 seconds.

## 3. How fast will the golf ball be traveling when it hits the ground?

The final velocity of the golf ball can be calculated using the equation v = u + at, where v is final velocity, u is initial velocity (0 m/s), a is acceleration due to gravity (9.8 m/s²), and t is time (8.8 seconds). The final velocity of the golf ball when it hits the ground will be approximately 86.24 m/s (193 mph).

## 4. Will air resistance affect the position and velocity of the golf ball?

Yes, air resistance will affect the position and velocity of the golf ball. As the ball falls, it will experience air resistance, which will decrease its acceleration and cause it to reach the ground at a slower velocity than it would without air resistance. However, since the Empire State Building is not very tall in comparison to the size of the Earth, the effect of air resistance will be minimal.

## 5. What factors can affect the position and velocity of the golf ball when dropped from the Empire State Building?

Aside from air resistance, other factors that can affect the position and velocity of the golf ball include the initial angle at which it is dropped, the mass and size of the ball, and external forces such as wind. These factors can alter the trajectory and speed of the ball, leading to variations in its final position and velocity when it reaches the ground.

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