Solving Projectile Motion: What Step Does Ball Hit First?

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A ball rolls off a stairway horizontally at 1.52 m/s, with each step measuring 1.52 cm in height and 20.3 cm in width. To determine which step the ball hits first, the projectile motion equations can be applied, specifically using the time of flight and horizontal distance traveled. The time of flight is calculated with T = √(2h/g), where h is the height of the steps and g is the acceleration due to gravity. The horizontal distance covered is given by nw = v/g * √(2h/g), where n is the number of steps passed and w is the width of each step. Ultimately, it's important to round up when calculating the number of steps down the ball travels.
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A ball rolls horizontally off the top of a stairway with a speed of 1.52m/s. The steps are 1.52 cm high, and 20.3 cm wide, which step does the ball hit first.

I don't really know where to start, I kept trying to use the formulas..but I would always end up with two unknowns.

Can anyone point me in the right direction?
 
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suspenc3 said:
A ball rolls horizontally off the top of a stairway with a speed of 1.52m/s. The steps are 1.52 cm high, and 20.3 cm wide, which step does the ball hit first.
I don't really know where to start, I kept trying to use the formulas..but I would always end up with two unknowns.
Can anyone point me in the right direction?
The range of the ball is vT, where v = 1.52m/s and T is the duration: T = ROOT(2gnh)/g where h is the height of a step and n is the number of them passed. The range is also the sum of the widths of the steps passed, so:

nw = v/g x ROOT(2gnh)

where w = 20.3 x 10^-2.

Rearrange and solve for n, rounding down.
 
Sorry, I was counting steps 'across' when I said round down. You should round up for how many steps down the ball goes. Makes more sense.
 
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