No variables involved!

Classical Mechanics Level 3

Jack, Jill, and John are on the roof of a building.

  • Jack drops a ball of mass 12 12 g g and the ball reaches the ground in time t j a c k = 2 s . t_{jack} = \SI{2}{\second}.
  • Jill throws a different ball of mass 10 10 g g upwards with some velocity and the ball falls to the ground in time t j i l l = 8 s t_{jill} = \SI{8}{\second} .
  • John throws a different ball of mass 16 16 g g downwards with exactly the same speed as Jill 's ball.

Find the time t j o h n t_{john} (in seconds) that John's ball takes to hit the ground.


The answer is 0.5.

Ayon Ghosh by Ayon Ghosh , 17, India

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1 solution

Steven Chase
Mar 29, 2017

The masses are irrelevant. Suppose the building height is h h . The Jack equation is:

− h = − 1 2 g t J a c k 2 − h = − 2 g \large{-h = - \frac{1}{2} g \, t_{Jack}^2 \\ -h = -2g}

The Jill equation is:

− h = v t J i l l − 1 2 g t J i l l 2 − h = 8 v − 32 g \large{-h = v \, t_{Jill} - \frac{1}{2} g \, t_{Jill}^2 \\ -h = 8v - 32g}

Equating the two gives:

30 g = 8 v ⟹ v g = 30 8 \large{30g = 8v \implies \frac{v}{g} = \frac{30}{8}}

When Jill's ball comes back down to the top of the building, it has the same speed it went up with, but now in the downward direction. This is the key to relating Jill and John (since that is John's initial velocity). The Jill time is equal to the John time plus twice the time it takes for gravity to sap away all of the initial velocity.

t J i l l = t J o h n + 2 v g 8 = t J o h n + 2 30 8 = t J o h n + 7.5 t J o h n = 1 2 \large{t_{Jill} = t_{John} + \frac{2v}{g} \\ 8 = t_{John} + 2 \frac{30}{8} = t_{John} +7.5 \\ t_{John} = \boxed{\frac{1}{2}} }

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Steven Chase very good solution.

Ayon Ghosh - 4 years, 2 months ago

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