# Physics help needed....Acceleration and distance **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [March 9, 2000, 11:44pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725 "2000-03-09T23:44:00Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![ChuckSki](https://avatars.discourse-cdn.com/v4/letter/c/51bf81/32.png) [@ChuckSki](https://boards.straightdope.com/u/ChuckSki)\ **Post date:** [March 10, 2000, 10:07pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/1 "2000-03-10T22:07:00Z")
I want to throw a softball a distance of 75 meters here on the surface of the Earth. Given a constant acceleration of 9.81m/sec/sec (or 32ft/sec/sec), what is the best angle at which to throw the ball to use the least amount of force, and what is the required initial acceleration? You may ignore the air resistance for this hypothetical, or include it if you so choose. I’ve selected a softball, but the (sans air resistance) answers should be the same for any body, right? I know the force required would be dependent on the weight of the object, but the acceleration required is the same, right? This is not an exercise in futility, as I do have a practical application for the results. Can anyone help? I’ve tried to do this myself, and have blown an afternoon doing so. If there’s a formula, I’d love to see it so that I can incorporate it into an Excel spreadsheet. ---
**Author:** ![Andy](https://avatars.discourse-cdn.com/v4/letter/a/aeb1de/32.png) [@Andy](https://boards.straightdope.com/u/Andy)\ **Post date:** [March 10, 2000, 10:13pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/2 "2000-03-10T22:13:00Z")
I’ll give you a clue, there are two solutions for theta when you solve for tan squared theta. Solve vertically and horizontally for forces and you’ll be there. ---
**Author:** ![ReservoirDog](https://avatars.discourse-cdn.com/v4/letter/r/8edcca/32.png) [@ReservoirDog](https://boards.straightdope.com/u/ReservoirDog)\ **Post date:** [March 10, 2000, 10:22pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/3 "2000-03-10T22:22:00Z")
Hey, hang on a second there, Chuck. We don’t do math over here. If we wanted to think, we would go to GQ. ---
**Author:** ![Strainger](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/strainger/32/244_2.png) [@Strainger](https://boards.straightdope.com/u/Strainger)\ **Post date:** [March 10, 2000, 10:23pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/4 "2000-03-10T22:23:00Z")
When I was a freshman in college, we did a proof that showed that 45 degrees was the angle to throw the object at to achieve the greatest distance (I don’t remember the proof off the top of my head and don’t have time to mess with it right now). That 45 degrees is _above_ the horizontal, of course. I don’t understand the acceleration part of the question. The greater the acceleration during the throw, the greater the initial velocity and the greater the distance. ---
**Author:** ![Michael\_Masterson](https://avatars.discourse-cdn.com/v4/letter/m/53a042/32.png) [@Michael\_Masterson](https://boards.straightdope.com/u/Michael_Masterson)\ **Post date:** [March 9, 2000, 11:44pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/5 "2000-03-09T23:44:00Z")
Hola, Chuckles, a firing angle of 45 degrees gives the maximum horizontal range–here’s why: R=[(v^2)(sin2x)]/g The max value for sin2x is 1, this happens when 2x=90 degrees and that would make x=45 degrees (pretty simple huh, if you want I can show you how I derived this formula, just ask). O yeah, R=horizontal range, v= initial velocity (you can think of it as speed in your case) and x=theta, or your initial angle. Tu comprendas? A deeper understanding as to why its 45 degrees is due to the principle of least action, but i’m tired of typing. * * * ~-MCM-~ ---
**Author:** ![Imthecowgodmoo](https://avatars.discourse-cdn.com/v4/letter/i/58f4c7/32.png) [@Imthecowgodmoo](https://boards.straightdope.com/u/Imthecowgodmoo)\ **Post date:** [March 10, 2000, 3:04am UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/6 "2000-03-10T03:04:00Z")
I’d throw my $.02 in, but I dropped Physics this semester. * * * “People must think it must be fun to be a super genuis, But they don’t realize how hard it is to put up with all the idiots in the world.” – Calvin and Hobbes (\_\_) _/_ ---
**Author:** ![Bricker](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@Bricker](https://boards.straightdope.com/u/Bricker)\ **Post date:** [March 10, 2000, 3:20am UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/7 "2000-03-10T03:20:00Z")
Credit where credit is due. Michael – excellent post! - Rick ---
**Author:** ![ChuckSki](https://avatars.discourse-cdn.com/v4/letter/c/51bf81/32.png) [@ChuckSki](https://boards.straightdope.com/u/ChuckSki)\ **Post date:** [March 13, 2000, 12:33pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/8 "2000-03-13T12:33:00Z")
Thanks to all. Good explanation, Michael, and sufficient to get me through my delema. Mucho gracias. Chuck L. * * * “The intellectuals’ chief cause of anguish are one another’s works.” Jacques Barzun Cheers! CAL ---
**Author:** ![dragonfly98](https://avatars.discourse-cdn.com/v4/letter/d/7bcc69/32.png) [@dragonfly98](https://boards.straightdope.com/u/dragonfly98)\ **Post date:** [March 15, 2000, 7:43pm UTC](https://boards.straightdope.com/t/physics-help-needed-acceleration-and-distance/14725/9 "2000-03-15T19:43:00Z")
> [@](#): > > Originally posted by Michael Masterson: > Hola, Chuckles, a firing angle of 45 degrees gives the maximum horizontal range–here’s why: \*\*R=[(v^2)(sin2x)]/g \*\*The max value for sin2x is 1, this happens when 2x=90 degrees and that would make x=45 degrees (pretty simple huh, if you want I can show you how I derived this formula, just ask). Alright Mike. I’m asking. It’s been a while for me. 🙂 * * * “What’s right is only half of what’s wrong and I want a short-haired girl who sometimes wears it twice as long.” George Harrison - _Old Brown Shoe_