Post a New Question

Physics

posted by .

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?

  • Physics -

    d = vt - horizontal distance = 0.544
    s = 1/2 at^2 - vertical distance = 1.19
    so,
    t = √(2s/a)
    v = d/t = d/√(2s/a) = .544/√(2*1.19/9.8) = 1.10m/s

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question