SOMEONE ASKED 👇
Steelhead trout migrate upriver to spawn.?
HERE THE ANSWERS 👇
the height h(t) = -4.9t^2 + 8t
h(t)=1.8 —> t = 0.3 or 1.4 seconds
Hmax = Voy^2/2g = 8^2/19.6 = 3.27 m
1.8 = 8*t-4.9*t^2
t = (8-√8^2-19.6*1.8)/9.8 = 0.27 sec
t = (8+√8^2-19.6*1.8)/9.8 = 1.36 sec
What’s the meaning of these two times?
With such an initial vertical speed (8.0 m/sec) , the steelhead reaches a top height of 3.27 m and then fall down , therefore the height of 1.8 m is met twice, the first time after 0.27 sec while ascending and the second while descending .
Since the motion can’t be stopped while ascending , it is reasonable to consider the longer time as the proper answer to such problem !!!