ball thrown vertically upwards physics?

With the displacement formula (x = v0 *t + (1/2) a t^2), you have all the values except the unknown initial velocity. So solve for that.

x = v0 t + (1/2) a t^2

v0 t = x - (1/2) a t^2

v0 = (x - .5 a t^2) / t

v0 = (40m - .5 (9.81m/s^2) (4s)^2) / 4s

v0 = (40m - .5 (9.81 m/s^2) (16s^2)) / 4s

v0 = (40m - 78.5m) / 4s

v0 = -38.5m / 4s

v0 = -9.6 m/s

Since gravitational acceleration is downward (and positive), the negative initial velocity implies an upward direction.

v = v0 + a t

v = -9.6m/s + 9.81m/s^2 * 4s

v = -9.6m/s + 39.24 m/s

v = 29.6 m/s

1/2 A T^2 = distance, so 3 sec = 45 m &(30 m/s) down, up at 10m/s should stop in one meter with 5 meter displ.