You are in an egg drop contest which you win by breaking your egg the hardest. You drop your egg off of a 3 story building (12m above the ground). You can throw it up with speed 11m/s, or straight down with speed 11m/s.

Question

How high above the ground will the upward thrown egg reach in m?

Answer 1

To solve this problem, we can use the kinematic equation:

h = h0 + v0t - 1/2gt^2

where:
h = final height (what we're solving for)
h0 = initial height (12m above ground)
v0 = initial velocity (11m/s upwards)
g = acceleration due to gravity (-9.8m/s^2)
t = time (what we're solving for)

We want to find the height reached by the egg when thrown upwards, so we'll set the vertical velocity to 0 at the maximum height (when the egg is at the top of its trajectory). This gives us:

0 = v0 - gt

t = v0/g

Plugging this in to the first equation, we get:

h = h0 + v0(v0/g) - 1/2g(v0/g)^2

h = 12 + 11(11/-9.8) - 1/2(-9.8)(11/-9.8)^2

h = 12 - 60.5 + 6.06

h = -42.44

Uh-oh! This means that the egg does not reach a positive height when thrown upwards, so it would break upon impact with the ground. It turns out that the initial velocity is not high enough to overcome the acceleration due to gravity and reach a positive height.

Therefore, the best strategy would be to drop the egg straight down from the building (with an initial velocity of 0) in order to break it the hardest.