Answer
The ball hits the third step.
Work Step by Step
We can find the time to fall the first $20.3~cm$:
$0.203~m = \frac{1}{2}a_yt^2$
$t = \sqrt{\frac{(2)(0.203~m)}{9.8~m/s^2}}$
$t = 0.20354~s$
We can find the horizontal distance the ball moves in this time:
$x = (1.52~m/s)(0.20354~s) = 0.309~m$
Since the horizontal distance is greater than the width of the first step, the ball misses the first step.
We can find the time to fall two steps of $20.3~cm$ each:
$0.406~m = \frac{1}{2}a_yt^2$
$t = \sqrt{\frac{(2)(0.406~m)}{9.8~m/s^2}}$
$t = 0.28785~s$
We can find the horizontal distance the ball moves in this time:
$x = (1.52~m/s)(0.28785~s) = 0.438~m$
Since the horizontal distance is greater than the width of the first two steps, the ball misses the first two steps.
We can find the time to fall three steps of $20.3~cm$ each:
$0.609~m = \frac{1}{2}a_yt^2$
$t = \sqrt{\frac{(2)(0.609~m)}{9.8~m/s^2}}$
$t = 0.35254~s$
We can find the horizontal distance the ball moves in this time:
$x = (1.52~m/s)(0.35254~s) = 0.536~m$
Since the horizontal distance is less than the width of the first three steps, the ball hits the third step.
