- #1
grayb
- 4
- 0
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!
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!