1) If we assume no friction, the horizontal velocity should be constant until the ball hits something. The vertical velocity should vary according to
-9.8m/(sec^2) or 9.8m/(sec^2) depending on which orientation is used for positive vertical velocity.
2) If the vertical velocity did not vary, the ball would hypothetically continue upward at an angle of 35 degrees as it passes the wall. This means that absent the downward acceleration, the ball would pass over the wall X meters higher than its starting height, where X/140 = tangent of 35 degrees, if I remember the definition of tangent correctly.
This means the no-gravity increase over its starting height at that point would =
140 * tan(35deg)
so the no-gravity clearance over the wall would be
(140 * tan(35deg)) - 24
3) The fact that the ball barely clears the wall means that the time it took the ball to travel from its starting point to the wall is just enough time for the ball lose
((140 * tan(35deg)) - 24) meters of height compared to the height the ball would have had with no gravity pulling down on it. I am pretty sure that should be the same amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
4) Let T = the amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
Since we are considering distance downward as a positive number and time as positive, for this portion we should consider acceleration caused by gravity to be
positive 9.8m/(sec^2) .
5) I am pretty sure for an object starting at rest
time = square root of ((2*distance)/acceleration)
-- so if I may take the liberty of representing square root as '^0.5', then T referred to in #4 above should be
6) I believe T calculated from #5 above is the time it takes for the ball to travel to the wall. The horizontal component of the ball's velocity is
140 divided by T meters-per-second
7) Picture a right-triangle with a horizontal leg 140 meters, and the hypotenuse angled upward at 35 degrees from that leg. The length of that hypotenuse =
8) I think the ball's initial velocity is
-- or, same value as above --
where T is the time it takes the ball to reach the wall, from #4, #5, and #6 above.
Assuming #5 above is correct and T actually is
then I think the ball's initial velocity should be