Answer:
Explanation:
vertical component of the velocity vy = 0 m/s
horizontal component of the velocity vx = 1.67 m/s
t, time for the water to reach the ground = √(2h / g) derived from the equation h = ut + 1/2 gt²
t = √ ( 2 × 3.70 / 9.8) = 0.869 s
a) the space left will = vx × t = 1.67 m/s × 0.869 s = 1.45 m
b) re scaling the height from which the water traveled = 3.70 m / 12 = 0.308 m
then the time for the travel in the model = √(2h/g) = √ ( 2×0.308 / 9.8) = 0.251 s
then the horizontal displacement re scaling = 1.45 m / 12 = 0.121 m
0.121 m = vx × 0.251 s
how fast should the water in the channel flow vx = 0.121 m / 0.251 s = 0.482 m/s