# Ball physics question

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1 hour ago, taby said:

I'm horrible at calculus. ﻿

Davean!

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Posted (edited)

Ok here's another question related but with different approach.

If ball has to rise to 10 meters in 1 second, what must the initial velocity be and what acceleration (So that ball drops back to ground at 2 seconds)?

Edited by heh65532

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Posted (edited)

If I'm not mistaken:

v(t) = v0 + a * t (with v0 the initial velocity and a the acceleration)
p(t) = v0 * t + (1/2) * a * t2 (the ball's vertical position)

At the top t = tr (r for "rise", 1 s in your case) v(tr) = 0 so v0 = -a * tr (eq 1)
p(tr) = h (10 meters) so h = v0 * tr + (1/2) * a * tr2 = - (1/2) * a * tr2 (using eq 1)

And finally:
a = -2 * h / tr2 = -20 m/s²
and
v0 = 2 * h / tr = +20 m/s

(the descent is symmetric and the ball will get down in 1 s and reach the ground at 2 s falling at -20 m/s)

Edited by D.C.Elington

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Posted (edited)
10 hours ago, D.C.Elington said:

If I'm not mistaken:

v(t) = v0 + a * t (with v0 the initial velocity and a the acceleration)
p(t) = v0 * t + (1/2) * a * t2 (the ball's vertical position)

At the top t = tr (r for "rise", 1 s in your case) v(tr) = 0 so v0 = -a * tr (eq 1)
p(tr) = h (10 meters) so h = v0 * tr + (1/2) * a * tr2 = - (1/2) * a * tr2 (using eq 1)

And finally:
a = -2 * h / tr2 = -20 m/s²
and
v0 = 2 * h / tr = +20 m/s

(the descent is symmetric and the ball will get down in 1 s and reach the ground at 2 s falling at -20 m/s)

Thanks! The first part is total mystery to me but even I can understand the final formula

So let me get this straight, for the final formula I input h = 10 and t = 1 and then I get the answer to my question?

Edit:

Actually I have little difficult to understand where the 2 and -2 comes in the last formula? Is that the time ball drops back to ground?

Edited by heh65532

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Posted (edited)
3 hours ago, heh65532 said:

for the final formula I input h = 10 and t = 1 and then I get the answer to my question?

Yes that's it

3 hours ago, heh65532 said:

Actually I have little difficult to understand where the 2 and -2 comes in the last formula?

This is equation manipulation. Starting from h = - (1/2) * a * tr2 and wanting to isolate 'a'...
=> multiply both sides by -2, this yields :
-2 * h = -2 * (-1/2) * a * tr2 and then -2 * h = a * tr2 because -2 * (-1/2) = -2 / -2 = 1
=> then divide both sides by
tr2 to finally obtain -2 * h / tr2= a

Now about the original (1/2):  position is the "accumulation of velocity" over time (that's why you get the famous "distance = velocity * time" when the velocity is constant). In maths terms this is called "integrating".
And precisely when integrating a * t over time with a constant (in v(t) = v0 + a * t), the result is (1/2) * a * t2... and that's really a calculus rule sorry

Edited by D.C.Elington

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Posted (edited)

@D.C.Elington Well I don't understand all that. My question is do I keep the 2 & -2 in the final formula always the same?

Edited by heh65532

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1 hour ago, heh65532 said:

@D.C.Elington Well I don't understand all that. My question is do I keep the 2 & -2 in the final formula always the same?

Ah yes sorry, they are part of the formula itself and not related to the specific numerical inputs in your example.

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16 minutes ago, D.C.Elington said:

Ah yes sorry, they are part of the formula itself and not related to the specific numerical inputs in your example.

Alright, good to know I don't have to worry about those  Got your math working in my test app. Thanks again!

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