- #1
mattst88
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Problem: A water tank develops a leak 20 meters below the surface of the water. The tank is 20 meters above the ground and the hole leaking water is 0.5cm in diameter. How far from the tank will the water hit the ground?
I used P1 + (1/2)(ρ)(v1^2) + ρgy = P2 + (1/2)(ρ)(v2^2) + ρgy
I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.
I assumed the tank was open topped and therefore P1 = P2.
We're left with:
ρgy = (1/2)(ρ)(v2^2)
so I canceled the ρ, leaving:
(9.8 m/s^2)(20 m) = (1/2)(v2^2)
so v = sqrt(98) at an angle of 0 degrees, right?
-------
yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground
if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 20 meters away from the tank.
Can someone verify my answer or find any errors in my logic or calculations? Thanks
I used P1 + (1/2)(ρ)(v1^2) + ρgy = P2 + (1/2)(ρ)(v2^2) + ρgy
I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.
I assumed the tank was open topped and therefore P1 = P2.
We're left with:
ρgy = (1/2)(ρ)(v2^2)
so I canceled the ρ, leaving:
(9.8 m/s^2)(20 m) = (1/2)(v2^2)
so v = sqrt(98) at an angle of 0 degrees, right?
-------
yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground
if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 20 meters away from the tank.
Can someone verify my answer or find any errors in my logic or calculations? Thanks