# Calculating Baseball Travel Time and Distance: 165 km/h, 18.3 m

• Kildars
In summary, the conversation discusses a problem involving a baseball being thrown horizontally at a speed of 165 km/h towards a batter who is 18.3 m away. The question asks for the time it takes for the ball to travel the first half and second half of the distance, as well as the distance the ball falls under gravity during the first half and second half. It is determined that the time for both halves is 0.1996 seconds and the distance the ball falls in the first half is 0.1952 m. The correct calculation for the distance the ball falls in the second half is to subtract the answer in C from the total distance, which gives a final result of 0.3904 m.
Kildars
A baseball leaves a pitcher's hand horizontally at a speed of 165 km/h. The distance to the batter is 18.3 m. Neglect air resistance.

(a) How long does it take for the ball to travel the first half of that distance?
s
(b) How long does it take for the ball to travel the second half of that distance?
s
(c) How far does the ball fall under gravity during the first half?
m
(d) How far does the ball fall under gravity during the second half?
m

For A I know D and V so I just did v = d/t which translates to t = d/v

It doesn't work.. What am I doing wrong?

Last edited:
I suggest you write down the equations for projectile motion. It might be useful to type 'baseball' or 'projectile motion' into the search box, since this is a typical kinematics problem which appears very often here. Anyway, present some of your work, and we'll be glad to help.

I suggest you write down the equations for projectile motion. It might be useful to type 'baseball' or 'projectile motion' into the search box, since this is a typical kinematics problem which appears very often here. Anyway, present some of your work, and we'll be glad to help.

:) I did t = d/v which is

Well ask for the time for the first half of the distance of the thrown ball..

So 18.3/2 = 9.15

then t = d/v
t = 9.15/165
Which = .05545 and that is wrong.

Oh I think I have to convert from km/h to m/s right?

Kildars said:
Oh I think I have to convert from km/h to m/s right?

Yes, of course. Watch out for the units.

Well I did it

165km/h = 45.833 m/s

so t = 9.15/45.833..Which gets you .1996 and that is still wrong.

You got the equation wrong; v = s / t => t = s / v.

t = s/v What's S?

Kildars said:
t = s/v What's S?

Oooops, I deleted my last post, because I wrote something stupid, please ignore it. I'm trying to figure this out right now.

Oh, alright.

So anyone?

Kildars said:
Well I did it

165km/h = 45.833 m/s

so t = 9.15/45.833..Which gets you .1996 and that is still wrong.
Except for the lack of units, this is not wrong. Did you state the problem correctly?

Nvm, that is right. I was just being stupid and typing it in the wrong box. So a and b are solved, I'm going to try C now.

So

A = .1996
B = .1996
C = .1952
D = ?

I tried doubling my time and just doing .5(-9.8)(.3992^2)

And got -.78086 Which is incorrect.

Kildars said:
So

A = .1996
B = .1996
C = .1952
D = ?

I tried doubling my time and just doing .5(-9.8)(.3992^2)

And got -.78086 Which is incorrect.

This would be correct if the question was 'How far did the ball fall during the whole time of its flight?'. You just have to subtract the answer in C from this.

Nice, thanks :).

Time for class now, I'll post more questions if i need help later.

## 1. How do you calculate the travel time for a baseball traveling at 165 km/h?

The formula to calculate travel time is distance divided by speed. In this case, the distance is 18.3 m and the speed is 165 km/h. First, convert 165 km/h to m/s by dividing by 3.6 (since 1 km/h is equal to 0.27778 m/s). This gives us a speed of 45.8 m/s. Then, divide the distance of 18.3 m by the speed of 45.8 m/s. The result is approximately 0.4 seconds.

## 2. What is the distance covered by a baseball traveling at 165 km/h for 0.4 seconds?

To calculate the distance, we use the same formula as above: distance = speed * time. In this case, the speed is still 165 km/h (converted to 45.8 m/s) and the time is 0.4 seconds. Multiplying these values gives us a distance of approximately 18.3 meters.

## 3. How does the speed of the baseball affect the travel time and distance?

The faster the speed, the shorter the travel time and the greater the distance covered. This is because the speed is directly proportional to both time and distance in the formula distance = speed * time.

## 4. Can the travel time and distance be calculated for a baseball thrown at a different speed?

Yes, the travel time and distance can be calculated for any given speed. Simply plug in the new speed and the distance of 18.3 m into the formula distance = speed * time and solve for time. The resulting time will be the travel time for the new speed. To calculate the distance, use the same formula but solve for distance instead.

## 5. How accurate is the calculation for baseball travel time and distance?

The calculation is accurate within a certain margin of error. This margin of error can be affected by various factors such as air resistance, wind, and the height from which the ball is thrown. However, for most practical purposes, this calculation provides a good estimate of the travel time and distance for a baseball thrown at a constant speed of 165 km/h.

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