This is a question I sometimes ask interview candidates.
To make it easier, you need to divide the problem up in two ways
- first, divide the arc into an upward and a downward portion
- then divide each part into a vertical and horizontal component
You need to use the equations (1) d = v0*t + 0.5*a*t^2 and (2) v1 = v0 + a*t.
The heights at A, B, and C are hA, hB, and hC. Then the height of the rising arc is hAC and the height of the falling arc is hCB.
Consider vertical component of the rising arc. Using equation (2), hAC = 0.5*g*t1^2, so t1 = sqrt(2*hAC/g).
You can solve for the initial vertical component of velocity using (2) to get sqrt(2*g*hAC).
Similarly for the falling arc, t2 = sqrt(2*hCB/g).
The total time is t1+t2.
If the horizontal distance is dAB, then the horizontal component of velocity (a = 0) from (1) is dAB/(t1+t2).
[EDIT] Fixed initial vertical velocity (was using hCB, now correctly using hAC). Oops.
To make it easier, you need to divide the problem up in two ways
- first, divide the arc into an upward and a downward portion
- then divide each part into a vertical and horizontal component
You need to use the equations (1) d = v0*t + 0.5*a*t^2 and (2) v1 = v0 + a*t.
The heights at A, B, and C are hA, hB, and hC. Then the height of the rising arc is hAC and the height of the falling arc is hCB.
Consider vertical component of the rising arc. Using equation (2), hAC = 0.5*g*t1^2, so t1 = sqrt(2*hAC/g).
You can solve for the initial vertical component of velocity using (2) to get sqrt(2*g*hAC).
Similarly for the falling arc, t2 = sqrt(2*hCB/g).
The total time is t1+t2.
If the horizontal distance is dAB, then the horizontal component of velocity (a = 0) from (1) is dAB/(t1+t2).
[EDIT] Fixed initial vertical velocity (was using hCB, now correctly using hAC). Oops.