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mybrohshi5
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Homework Statement
Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 27.5 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation Θ of the hose until the water takes 3.00 s to reach a building 51.81 m away. You can ignore air resistance; assume that the end of the hose is at ground level.
How fast is it moving just before it hits the building?
Θ=51.1
Homework Equations
The equation i used to solve it was V=sqrt(vx2+vy2) and the answer was found to be 19.0 m/s
i was wondering if this equation would work to solve it? i can't get it to come out with 19 m/s though.
Vf2= Vyi2+2(ay)(Δy)
The Attempt at a Solution
Vf2= Vyi2+2(ay)(Δy)
so with this equation i used Vyi to be 27.5sin(51.1)=21.01
i used ay to be -9.8
i used Δy to be 27.5sin(51.1)(3) + .5(-9.8)(32) = 20.11 - 0 = 20.11
(the minus 0 is because of the final height minus the initial height).Vf2= Vyi2+2(ay)(Δy)
Vf = 6.87 which is definitely wrong.
Anyone know why this equation doesn't work using these numbers i calculated?
Any suggestions. Thanks :)