- #1

mattst88

- 29

- 0

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?

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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