Here's how I would do it:
Let's disregard the first part of the projectile (from the time when you throw it up until it reaches its maximum height).
Let's just look at the second half of the projectile- the point where it reaches its maximum height until it lands in your hands again.
Using the equation y = v(initial)t + 1/2 a t^2: when it's at its maximum height, its vertical acceleration is 0, so v(initial) = 0
So the above equation becomes: y = 0 + 1/2 at^2.
From that equation, we know that t ~ square root (1/a) , in other words, t is inversely proportional to the sqrt(a) (just like what you had said, sorry-i had to work it through).
Okay, so using
t~1/ sqrt a:we know that a(earth) = 2*a(moon).
---> time for earth = t~ 1/sqrt a
--> time for moon = t' ~ 1/ sqrt (1/.5a) --> because the acceleration on the moon is half that of earth
finally, t' = 1/ sqrt (2/a) --> because 1/.5 = 2
So on the moon, it would take sqrt 2 times longer than that of the Earth
The time that we just calculated is the SECOND half of the projectile, which means that it's only half of the whole time. However, I don't think it matters because this is a ratio problem, not an absolute answer problem.
Sorry for the confusing wording, I know my answer and Bernoull's are different, and honestly, if someone could point out why is there a discrepancy, I would appreciate it.
JohnDoe: what does the solutions say? I'm curious.
