The Physics Behind The Cannonball Scene: Home Alone 2
In case you've forgotten the scene...
https://youtu.be/QhePJCracho?t=2296
Link to fast forward to scene^

Where's the physics?

From this scene, we can determine that Kevin demonstrates projectile motion when he launches himself into the air at a certain degree in order to cannonball effectively into the pool. We can assume the angle of his jump being 50 degrees and the length of his jump as 2 metres, with the jump lasting 1 second long.  With this information we can find the depth of the pool he jumps into in metres. 

Finding the depth of the pool: 

x stuff:
△dx = 2m

vx = 1Cos50 = 0.64m/s

t = 1s

y stuff:

vy1 = 1Sin50 = 0.77m/s

a = -9.8 m/s^2

t = 1s

△dy = ?

vy2 = ?

△d = v1t + 1/2at^2

      = (0.77) (1) + (-4.9) (1)^2

      = -4.13

∴ h = 4m

Now that we have a value for the depth of the pool, we can find the maximum height Kevin jumps... 

△dy = ?

vy1 = 0.77m/s

vy2 = Ø

a = -9.8m/s^2

vs = v1^2 +2a△d

0 = (0.77)^2 + 2(-9.8)(△d)

   = 0.5929 -4.9△d

-0.5929/-4.9 = △d-4.9/-4.9

0.121 = △d

∴ △d = 0.12m