At first it seems we have too little information, but it turns out the angle is invariant with the size of the larger square — grab point H and drag it left and right:

(GeoGebra resource by John Golden.)

The ball is always working against air resistance, so it’s losing energy all the time; at a given height, its total energy while rising is greater than while falling. The ball’s potential energy at this height is the same in both cases, so its kinetic energy (and hence its speed) is greater when it’s rising. So it takes longer to fall than to rise.

1. Kf5! Now the bishop must move, and then 2. h8=Q+ Bb8 3. Qh1#

Note that the king’s waiting move must be to f5. Moving to a black square leaves him vulnerable to a check; moving to the h-file blocks the queen’s access to h1; and moving to f3 blocks the queen’s final checkmate.

If 1. Kf5 Be5 then 2. h8=Q+ Bxh8 3. f8=Q#.

From Eskilstuna-Kuriren, 1931.

“If coal is burnt on the third floor height, the potential energy of the combustion products (water, ashes, carbon dioxide, carbon monoxide, and unburnt coal particles) will increase exactly as much as the coal potential energy.”

From Lange’s 1987 book Physical Paradoxes and Sophisms.

If all four equations are satisfied, then multiplying all four equations together will produce another valid equation. We obtain:

a⁷ × b⁶ × c⁵ × d⁴ × e³ × f² × g = 2,677,277,333,530,800,000.

The number on the right factors into prime powers as 2,677,277,333,530,800,000 = 2⁷ × 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17, as can be determined by trial division by the small primes 2, 3, 5, 7, 11, 13, and 17, for instance. So we have the new equation:

a⁷ × b⁶ × c⁵ × d⁴ × e³ × f² × g = 2⁷ × 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17.

If p is any prime dividing a, then p⁷ divides the right hand side. Since only 2 appears to the seventh power, the only value p can take is 2. Since a must be greater than 1, we see that 2 must divide a. Because 2 appears to exactly the seventh power on the right, we see that a = 2. Canceling a⁷ on the left with 2⁷ on the right leaves:

b⁶ × c⁵ × d⁴ × e³ × f² × g = 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17.

Repeating the same argument shows that b = 3, c = 5, d = 7, e = 11, f = 13, g = 17 is the unique solution to the new equation. However, it is still necessary that these values actually satisfy the original four equations, which in fact they do as can be checked easily. (To see why this last step is necessary, try using the same reasoning on the equations: x = 10, y² = 9, z³ = 25, which clearly has no solutions in integers.)