Asked by raquel

A sprinkler mounted on the ground sends out a jet of water at a 30 ∘ angle to the horizontal. The water leaves the nozzle at a speed of 11 m/s . How far does the water travel before it hits the ground?
Answered by Henry
Dx = Vo^2*sin(2A)/g.
Vo = 11 m/s.
A = 30o
g = 9.8 m/s^2.
Dx = ?
Answered by Marques
10.69
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