A sprinkler mounted on the ground sends out a jet of water at a 30o angle to the horizontal. The water leaves the nozzle at a sp
eed of 12 m/s. (a) How far does the water travel before it hits the ground?
1 answer:
Answer:
12.7m
Step-by-step explanation:
The equation for motion to be used to find how far the water travels is
xf = xi + Vix(t) + 1/2axt^2
xf = 0 + ViCos α(t) + 0
xf = ViCosα(t)
xf = 12 Cos 30°(t)
xf = (6√3)t
To find the total times we will use the vertical displacement equation
yf = yi + Viy(t) + 1/2ayt^2
0 = 0 + ViSinα(t) + (1/2 * (-9.8)t^2)
0 = 12Sin30°(t) - 4.9t^2
4.9t^2 = 12Sin30°(t)
4.9t^2 = 6t
Divide through by t
4.9t = 6
t = 6/4.9
t = 1.22secs
Put t = 1.22 into (6√3)t
xf = 6√3 * 1.22
xf = 12.7m
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