A small steel ball is projected horizontally off the top of a long flight of stairs. The initial speed of the ball is 3.4 m/s. Each step is 0.19 m high and 0.29 m wide. Which step does the ball strike first?

I am going to work this problem in two parts. First, I am going to only deal with vertical components to find out how much time it takes the ball to reach the first step:

Knowns:

a = g = -9.81 m/s^{2}

v_{i}= 0 m/s

x_{i} = 0 m

x_{f} = -.19 m

1/2at^{2} + v_{i}t + x_{i} = x_{f}

1/2(-9.81)t^{2} + 0t + 0 = -.19

t^{2} = 2*(-.19)/(-9.81)

t = .0387 s

Now, we will determine how far the ball traveled horizontally in this time.

Known:

v = 3.4 m/s

t = .0387 s

v = x/t

x = vt

x = 3.4*(.0387)

x = .132 m

Therefor, the ball strikes the top step. It would have to travel .29 m horizontally before dropping .19 m vertically in order to not hit the top step.