1 hour ago, taby said:

I'm horrible at calculus. 

https://xkcd.com/2117/

Davean!

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)?

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 = t_r (r for "rise", 1 s in your case) v(t_r) = 0 so v₀ = -a * t_r (eq 1)
p(t_r) = h (10 meters) so h = v₀ * t_r + (1/2) * a * t_r² = - (1/2) * a * t_r² (using eq 1)

And finally:
a = -2 * h / t_r² = -20 m/s²
and
v₀ = 2 * h / t_r = +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)

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 = t_r (r for "rise", 1 s in your case) v(t_r) = 0 so v₀ = -a * t_r (eq 1)
p(t_r) = h (10 meters) so h = v₀ * t_r + (1/2) * a * t_r² = - (1/2) * a * t_r² (using eq 1)

And finally:
a = -2 * h / t_r² = -20 m/s²
and
v₀ = 2 * h / t_r = +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?

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 * t_r² and wanting to isolate 'a'...
=> multiply both sides by -2, this yields :
-2 * h = -2 * (-1/2) * a * t_r² and then -2 * h = a * t_r² because -2 * (-1/2) = -2 / -2 = 1
=> then divide both sides by t_r² to finally obtain -2 * h / t_r²= 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   

@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?

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.

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!