old faithful geyser in yellowstone park shoots water hourly to a height of 44.8m. with what velocity does the water move when it is at 24.4m above the ground (on its way up)?
v(h)^2=vi^2 + 2g h
To find the velocity of the water when it is at 24.4m above the ground, we can use the principles of projectile motion. Specifically, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
s = displacement
a = acceleration due to gravity (-9.8 m/s^2)
In this case:
u = 0 m/s (since the water starts from rest)
s = 24.4m
We need to find v, so let's rearrange the equation:
v^2 = 0^2 + 2 * (-9.8) * 24.4
Simplifying:
v^2 = -2 * 9.8 * 24.4
v^2 = -478.4
Now, solve for v by taking the square root of both sides (since velocity can't be negative):
v ≈ √(-478.4)
v ≈ 21.9 m/s
Therefore, when the water is at 24.4m above the ground on its way up, it has a velocity of approximately 21.9 m/s.