在同一高度，在同一點，以相同速率v同時仰角x及俯角x度分別拋出,若兩者著地時間相差4秒且地點相距60m，則x為多少度?

Let R1 and t1 be the range and time of flight for the first object, R2 and t2 be the range and time of flight for the second object.
Let u be the initial projection velocity.

For the 1st object, we have: t1 = R1/(u.cos(x)) For the second object: t2 = R2/(u.cos(x))
Then, t1 - t2 = 4 But (t1-t2) = (R1-R2)/(u.cos(x)). This gives: 4 = 60/(u.cos(x)), i.e. u.cos(x) = 15
Consider the vertical motion: we have: for 1st object, -h = (u.sin(x))(t1) - (g/2)(t1)^2, where h is the height the object is projected. i.e. -h = (u.sin(x))(R1/(u.cos(x)) - (g/2)(R1/(u.cos(x)))^2
or -h = (u.sin(x))(R1/15) - (g/2)(R1)^2/225

Similarly for the 2nd object: we have:
-h = (-u.sin(x))(R2/15) - (g/2)(R2^2/225)
Equating the two equations:
(u.sin(x))(R1/15) - (g/2)(R1)^2/225 = (-u.sin(x))(R2/15) - (g/2)(R2^2/225) After simplifying, [(u.sin(x))/15].(R1+R2) = (g/450)(R1^2 - R2^2)
i.e. u.sin(x)(R1+R2) = (g/30)(R1+R2).(R1-R2)
or u.sin(x) = (g/30)(60)
i.e. u.sin(x) = 2g = 2 x 9.81 = 19.62
hence, (u.sin(x))/(u.cos(x)) = 19.62/15
i.e. tan(x) = 1.308
hence, x = 52.6 degrees