posted by Jackie on .
A jet of water squirts out horizontally from a hole near the bottom of the tank,
Assume that y = 1.19 m and x = 0.544 m. What is the speed of the water coming out of the hole?
d = vt - horizontal distance = 0.544
s = 1/2 at^2 - vertical distance = 1.19
t = √(2s/a)
v = d/t = d/√(2s/a) = .544/√(2*1.19/9.8) = 1.10m/s