a fish leaps vertically up towards the surface of water with speed 9m/s the height h in meters of the fish above the surface of the height h=9s-4.9s^2.approximently how many seconds after the fish lea[s will it hit the surface of the lake
Answers
Given : the height h in meters of the fish above the surface of the height h=9s-4.9s²
To Find : .approximately how many seconds after the fish leave will it hit the surface of the lake
Solution:
h=9s-4.9s²
h = 0 initially and again when it touches back surface
s = time
9s-4.9s² = 0
=> s(9 - 4.9s) = 0
=> s = 0 or s = 9/4.9
Fish hit the surface of the lake after 9/4.9 - 0
= 9/4.9 sec
= 90/49
= 1.8367
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Answer:
h=9s-4.9s²
h = 0 initially and again when it touches back surface
s = time
9s-4.9s² = 0
=> s(9 - 4.9s) = 0
=> s = 0 or s = 9/4.9
Fish hit the surface of the lake after 9/4.9 - 0
= 9/4.9 sec
= 90/49
= 1.8367
Step-by-step explanation: