Question

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 speed of 12 m/s. (a) How far does the water travel before it hits the ground?

Answer 1

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