Recall that the range
R = (v^2 sin2θ)/g
so, just plug in your numbers.
A sprinkler mounted on the ground sends out a jet of water at a 35∘ angle to the horizontal. The water leaves the nozzle at a speed of 15m/s . How far does the water travel before it hits the ground?
5 answers
13.17 m
0.22
7.85
V= 15m/s at 35° Vox = 15cos35 = 12.29 Voy = 15sin35° = 8.60
Ax = 0 bc acceleration is constant Ay = gravity = -9.8
Equation for time to max height is Vfy = Voy + at
0 = 8.6 + (-9.8)t
-8.6 = -9.8t
.8776s = t to max height
.8776 (2) = 1.755s = t
(.8776 was only the time to max height so double for full time.)
Distance traveled / displacement = Δx
Δx = Vox(t)
Δx = 12.29(1.755)
Δx = 21.57m
The water travels 21.6 m before hitting the ground.
Ax = 0 bc acceleration is constant Ay = gravity = -9.8
Equation for time to max height is Vfy = Voy + at
0 = 8.6 + (-9.8)t
-8.6 = -9.8t
.8776s = t to max height
.8776 (2) = 1.755s = t
(.8776 was only the time to max height so double for full time.)
Distance traveled / displacement = Δx
Δx = Vox(t)
Δx = 12.29(1.755)
Δx = 21.57m
The water travels 21.6 m before hitting the ground.
2 answers