World Record Longest Home Run: Roy "Dizzy" Carlyle

  • Thread starter Thread starter Motorboar
  • Start date Start date
  • Tags Tags
    Home
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 28K views
Motorboar
Messages
6
Reaction score
1

Homework Statement


According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-league game. The ball traveled 188m (618 ft.) before landing on the ground outside the ballpark.

Assuming that the ball's initial velocity was 40.0 degrees above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.900 m (3.00ft ) above ground level? Assume that the ground was perfectly flat.

How far would the ball be above a fence 3.00 m (10.0 ft) in height if the fence were 116 m (380ft ) from home plate?

Homework Equations


v0x=v0cos(40)
v0y=v0sin(40)
x-x0=v0xt
y-y0=v0yt+1/2at^2
ay=-o.81
y-y0=-0.9
ax=0
x-x0=188



The Attempt at a Solution


-0.9 =v0sin40t + 1/2(-9.81)t^2
t= 188/v0cos(40)
0 = (v0sin40)(188/v0cos(40))-(4.9)(188/v0cos(40))^2+0.9
graphed it on a calculator to avoid messy algebra and got the wrong answer of 32.97.
No Idea why this is wrong. Any ideas?
 
on Phys.org
Motorboar said:

Homework Statement


According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-league game. The ball traveled 188m (618 ft.) before landing on the ground outside the ballpark.

Assuming that the ball's initial velocity was 40.0 degrees above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.900 m (3.00ft ) above ground level? Assume that the ground was perfectly flat.

How far would the ball be above a fence 3.00 m (10.0 ft) in height if the fence were 116 m (380ft ) from home plate?

Homework Equations


v0x=v0cos(40)
v0y=v0sin(40)
x-x0=v0xt
y-y0=v0yt+1/2at^2
ay=-o.81
y-y0=-0.9
ax=0
x-x0=188



The Attempt at a Solution


-0.9 =v0sin40t + 1/2(-9.81)t^2
t= 188/v0cos(40)
0 = (v0sin40)(188/v0cos(40))-(4.9)(188/v0cos(40))^2+0.9
graphed it on a calculator to avoid messy algebra and got the wrong answer of 32.97.
No Idea why this is wrong. Any ideas?

I got 43.1 m/s... I just went with your numbers. I would try it again.
 
43.1 is the right answer. Did you graph it? I been working with the input for like 30 minutes, do you mind telling me how you plugged it in?
 
Motorboar said:
43.1 is the right answer. Did you graph it? I been working with the input for like 30 minutes, do you mind telling me how you plugged it in?

I just used your numbers. Maybe you squared the term for acceleration incorrectly?

-0.9 = sin40*(188/cos40) - 4.9*(188/Vo*cos40)^2

The Vo cancels out of the first term and the Vo in the second term gets squared so you eventually have to take the square root. It a bit of a pain as you said.
 
If I am not lazy and work it out by hand, I get it. Thanks for your help.