1. #1

    Help with a quick math problem!

    Solved. Thanks
    Last edited by Snakeseye0; 2014-10-05 at 03:02 AM.

  2. #2
    I don't know how you deal with those problems in US, but I would convert them to metrical units as it's much more practical to deal with any science related problem. I won't bother to give answer for you, but as hint: v = s/t so t = s/v

  3. #3

  4. #4
    Bloodsail Admiral Zvinny's Avatar
    10+ Year Old Account
    Join Date
    Dec 2013
    Location
    Detroit
    Posts
    1,228
    .46

    Google told me it's going 475,200 feet per hour, 7920 fpm or 132 fps or approximately 0.46 seconds to reach home plate.

    There ya go.

  5. #5

  6. #6
    Deleted
    Distance divided by velocity gives you the time.

  7. #7
    The Lightbringer Bluesftw's Avatar
    10+ Year Old Account
    Join Date
    Mar 2012
    Location
    Right here, right now
    Posts
    3,134
    if it was in metric it would be one simple equation / proportion and i assume same rules works in imperial so (0.011miles * 3600s)/90miles = 0,44 s , might be wrong tho^^

  8. #8
    The Unstoppable Force THE Bigzoman's Avatar
    10+ Year Old Account
    Join Date
    Mar 2012
    Location
    Magnolia
    Posts
    20,767
    Speed=Distance/Time.

  9. #9
    The distance will actually be a little bit more than 60 feet 6 inches since you're sort of pitching the ball downwards so the real distance will be:
    d = sqrt(18.4404^2 + h^2) where h is the height difference from where it's leaving the pitcher's hand and when it passes home plate.

  10. #10
    Quote Originally Posted by Snakeseye0 View Post
    If a major league pitcher throws a 90 mile per hour fastball, how many seconds does it take to reach home plate? The distance from the pitching rubber to home plate is 60 feet 6 inches.
    Nobody is taking aerodynamics into consideration here, we need to calculate the air resistance and spin of the ball while also taking gravity into consideration if we are to begin to work towards an accurate calculation.
    Most people would rather die than think, and most people do. -Bertrand Russell
    Before the camps, I regarded the existence of nationality as something that shouldn’t be noticed - nationality did not really exist, only humanity. But in the camps one learns: if you belong to a successful nation you are protected and you survive. If you are part of universal humanity - too bad for you -Aleksandr Solzhenitsyn

  11. #11
    Bloodsail Admiral Zvinny's Avatar
    10+ Year Old Account
    Join Date
    Dec 2013
    Location
    Detroit
    Posts
    1,228
    Quote Originally Posted by Venant View Post
    Nobody is taking aerodynamics into consideration here, we need to calculate the air resistance and spin of the ball while also taking gravity into consideration if we are to begin to work towards an accurate calculation.
    Not to mention the simple fact that a pitcher throws the ball at the full extent of his arm, meaning the distance the ball travles to the plate is a bit shorter than the actual distance from the mound to the plate

  12. #12
    Quote Originally Posted by Overdispersion View Post
    The distance will actually be a little bit more than 60 feet 6 inches since you're sort of pitching the ball downwards so the real distance will be:
    d = sqrt(18.4404^2 + h^2) where h is the height difference from where it's leaving the pitcher's hand and when it passes home plate.
    You can pretty safely assume the distance is close to the 18.4 metres. Even if it fell whole metre (which it doesn't), it would only add 2 cm to the total distance. That's pretty insignificant as ball would cross that distance in 0.5 ms. At half a metre drop it woudl only be 6 mm and that would be even more insignificant.

  13. #13
    At least 2 people attempted to answer the question.

    If you need a more technical, or less technical, explanation of anything below, please let me know.
    -----------------------------------

    This is mostly a unit conversion problem once you know what formula to use. I assume you have taken algebra and that you know the equation: (Distance) = (Rate) x (Time). D=RT. Instead of Rate, you might use Velocity (D=VT) or Speed (D=ST). I'll use R in my explanation.
    Since this problem is asking for Time and not Distance, the equation becomes: (Time) = (Distance)/(Rate). T=D/R.

    So let's deal with all of the units. The ultimate goal is to get them to all become the same units. We're dealing with miles, feet and inches. And we're dealing with hours and seconds. I think the most logical thing to do is convert everything into inches or seconds.

    Step 1.
    We start with 90 miles per hour. (Note: The exact way you'd know how to do this might be different than how I'll show it. Hopefully this makes sense.) I prefer fractional notation for these kinds of problems.

    90 mph = 90 miles/1 hour.

    I know there are 5280 feet in a mile (5280 ft = 1 mile), and there are 12 inches in a foot (12 in = 1 ft).

    To convert from one unit to the next, I multiply by a fraction that uses the conversion ratio for those units. (A conversion ratio is just two equal measurements written as a fraction, like 4quarts/1gallon, 1yard/3feet, 1foot/12inches. Since they're all the same measurement (4 quarts is the same as 1 gallon) but using different units, they balance out.) For example, to convert from miles to feet, I need the fraction 5280ft/1mile. The conversion ratio to go from feet to inches is 12in/1ft.

    Multiply the problem by its conversion ratio to go from one unit to the next. So to go from miles to feet:

    90miles/1hour * 5280ft/1mile = 475200ft/1hour.
    In this case, miles cancel out and feet are left over. (This is because miles is on the top of the first fraction and on the bottom of the second. When you divide something by itself, it cancels out.)

    Now we go from feet to inches:

    475200ft/1hour * 12inches/1ft = 5702400in/1hour.

    So, 90 mi/1 hr = 5702400 in/1 hr

    Step 2.
    Let's finish up step 1 by changing hours into seconds.

    There are 3600 seconds in an hour (3600 s = 1 hr). (This could also be done using 60 minutes = 1 hour and 60 seconds = 1 minute. It's longer that way, but perhaps more familiar.)

    5702400in/hr * 1hr/3600s = 1584 inches / 1 second.
    Note, in this case, the conversion ratio I used was 1hr/3600s. I put hours on top and seconds on bottom. I did that because the first fraction has hours on bottom, and to get rid of that unit I needed hours on top of the second fraction. If I had flipped the ratio (and use 3600s/1hr), I wouldn't have gotten rid of hours.

    So, 90 mi/1 hr = 1584 in/1 s.

    Step 3.
    Next, we'll deal with 60 feet and 6 inches. It would be best if this was all written in the same unit, so we'll change it to inches.

    Again, there are 12 inches in a foot.
    60ft * 12in/1ft = 720 in

    Add in the remaining 6 inches and you get 726 inches.

    So, 60 feet and 6 inches = 726 in.

    Step 4.
    Now we use the equation to calculate Time when we know Distance and Rate. T = D/R.

    In this problem,
    1. Distance was 60 feet and 6 inches. For that we got D = 726 in.
    2. Rate is 90 miles per hour. For that we got R = 1584 in / 1 s.

    Divide these two:
    726in / (1584in/1s) = 0.458 seconds.
    Last edited by Dendrek; 2014-10-04 at 10:23 PM.

  14. #14
    Deleted
    Five star post.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •