What is a quick way to approximate flight time off a cliff?

[grayb]

Hi, I'm basically trying to figure out if there's a quick way to approximate the flight time of something thrown downwards off a cliff.

If you throw a projectile off a cliff with height H > 0 above the ground, there are 3 cases:

v_y > 0
v_y = 0 (thrown horizontally off the cliff)
v_y < 0

For v_y > 0, it takes about v_y/10 seconds = t1 to reach the peak above the cliff and the new height is about H + (v_y)^2/20. Since v_y = 0 at the peak, it's like a freefall question from height H + (v_y)^2/20, so you can set H + (v_y)^2/20 = 5t^2, t2 = sqrt(H+(v_y)^2/20) and the total flight time is about t1 + t2.

If v_y = 0 then it's free fall and H = 5t^2 so t = sqrt(H/5).

For v_y < 0 (thrown downwards off a cliff), how can I approximate it without doing a quadratic equation? I guessed t1 + t2 - 1 but I'm not sure if that's right.

Thanks!

 

[TumblingDice]

This site has several pages related to gravity equations, including this one:
http://www.school-for-champions.com/science/gravity_equations_falling_time.htm

It looks to have calcs for all you describe, including initial velocity. (e.g., "thrown").

 

[CWatters]

If you assume constant acceleration the SUVAT equation..

s = ut + 0.5at²

(where u is the vertical velocity)

applies apply to all three cases.

One approximation you could make is to assume u is small.