I've been working on this problem for awhile but couln't get right result. So I hope anybody help me.

Lets say there's a man at height H. And he jumps stright upward and reaches maxH and fall back to ground(zero height) in given time(t).

What would be initial velocity(jump power) and gravity?

The man must reach to maxH(not over it) and fall back to ground in GIVEN TIME!

If he jumps from ground, it's solved by assuming that v=0 at half of given time.

But I can't solve if that man jumps from specific height.

Below is what I want to do.

void jump( float current_height, float max_height, float desired_time )

{

g = ...?

v0 = ...?

}

Thanks in advance.

I believe this to be a solution:

First, obtain Initial Velocity:

Initial vertical velocity as a function: Vb(Td, Ha, Hb ) = (-Td - ( Td^2 - Td^2/(Ha-Hb)*Ha )^(1/2) ) / ( Td^2 / (2*(Ha-Hb) ) )

Or in equation format:

tD = total time

HA = starting height

HB = max height

Once you have that, gravity can be derived:

Here's my derivation

Picture Link:

Document Link:

https://docs.google.com/file/d/0B39w-yzqjMIaMDNvT0RFczFHVlU/edit?usp=sharing ( google docs messes up the equations, download the docx to view them )

You can enter the above equation into any standard calculator that supports equations. (win power calc is what I used).

Testing it seems to work. For example:

Note that there may be simpler solutions, but I couldn't find any

Ooops, I forgot that there can be an angle for the jump

Angle? Angle has nothing to do with it, as far as I understood.

Sanghlee was only asking for vertical velocity (he mentioned a man jumping straight up).

More importantly, the calculations for vertical velocity can be entirely independently done from those for horizontal, because horizontal velocity has no effect on the time it takes an object to fall down.

SUPER THANKS to Milcho! Especially good derivation and super kind explanation!

I've not read details of your derivation yet but your fomula works perfectly!

You're right about the angle. I just needed upward force.

I'm impressed that you've gave excellent explanation to just a online question.

I'll be asking you if I have question while reading your derivation.

You helped me alot. Thanks again!

I believe this to be a solution:

First, obtain Initial Velocity:

Initial vertical velocity as a function: Vb(Td, Ha, Hb ) = (-Td - ( Td^2 - Td^2/(Ha-Hb)*Ha )^(1/2) ) / ( Td^2 / (2*(Ha-Hb) ) )

Or in equation format:

tD = total time

HA = starting height

HB = max height

Once you have that, gravity can be derived:

Gravity as a function: g( Td, Ha, V ) = (-Ha-V*Td)/(Td*Td)

Or in equation format:

tD = total time

HA = starting height

V = initial velocity, obtained from first equation

Here's my derivation

Picture Link:

Document Link:

https://docs.google.com/file/d/0B39w-yzqjMIaMDNvT0RFczFHVlU/edit?usp=sharing ( google docs messes up the equations, download the docx to view them )

You can enter the above equation into any standard calculator that supports equations. (win power calc is what I used).

Testing it seems to work. For example:

Vb(1.02, 0, 1.276) = 5.003

Vb(1.328, 2, 3.276) = 5.0008

You can check those values at http://www.had2know.com/academics/trajectory-parabola-equations-calculator.html - entering the corresponding values, and 90 for the "trajectory angle"

Note: Throughout my calculations I assumed the formulas looked like this: h = h0 + v0*t + g*t0^2 - meaning, I was adding gravity, instead of doing -g. Basically you have to realize that I was expecting the value of g to be negative. This will come into account when deriving gravity.

Edit 5: Added derivation work and visual formulas.

Note that there may be simpler solutions, but I couldn't find any

Glad to help - but as I said, there are probably simpler ways to calcualate the V0.

And here is one simplification: (to clarify, this is the exact same formula from above, only simplified)

V(Td, Ha, Hb) = 2*( Hb - Ha + (Hb*Hb - Ha*Hb)^(1/2) ) / Td

Same variable name/meanings - and it should run faster, and works just as well.