Find the step on which the hockey puck will land

[NotInMrPutmansClass]

Homework Statement

Find the step

Relevant Equations

0.2 stands for 20 cm (0.2m)

The question goes like this.
A frictionless puck is slid horizontally off the top of an infinite set of stairs with a speed of 3m/s. The steps are standard size: 20cm high and 20 cm wide. Which step will the puck strike first?Since the pluck will not get faster on the x-axis, these are my givens:

20 cm = 0.2mX-Axis:
Velocity = 3m/s
Distance = (0.2)X
Time = Y

Y-Axis
Int. Velocity = 0 m/s
Final Velocity = Unknown
Acceleration = 9.8 m/s^2
Distance = (0.2)X
Time = Y

Isolate for X in both equations

d=vt
(0.2)X=3Y
X= 3Y/0.2d=Init.V*t + 1/2 at^2
(0.2)X= 4.9Y^2
X = 4.9Y^2/0.2

Sub one of the X into the other formula

3Y/0.2 =3Y+4.9Y^2/0.2

After Isolating for Y (Using the Quadratic Formula) We get:

0= 0.98Y^2-0.6

Y1=0
Y2=0.612244 Seconds

now we sub this as the Y value in one of the formulas:

X(0.20) = 3Y
X(0.20) = 3(0.612244)
X = 9.1
The pluck will land on the 9th step.

But, if the steps are 20 cm wide. That means that if it lands on the 9th step, that would be the edge (because of the .1), so would it be the 10th step?

 

[kuruman]

I have not run the numbers, but if your calculation is correct and 9.1 is the answer, then yes, the puck will land on the 10th step. I will check the numbers and come back unless someone else does it first. If I were you, I would do the calculation symbolically and substitute at the very end. Here, roundoff errors matter if the final calculation is based on intermediate numerical calculations.

 

[kuruman]

I checked and I got the same answer as you. You're good to go.

One more thing:
I know that you can choose any symbols you want but to avoid confusion, in projectile motion we have reserved symbols
t = clock time (not Y)
x = horizontal distance at time t
y = vertical distance at time t.

Please conform to these if you don't mind. Thanks.

 

[haruspex]

the top of an infinite set of stairs

I would say that means it starts on the first step. If it were to start with a very small nonzero velocity it would land on the second step, etc. But my friend disagrees.

 

[PeroK]

If we take $y$ increasing downwards, then we just need to look for the point where the trajectory crosses the line $y = x$:$$y = \frac 1 2 gt^2, \ x = vt$$$$y = x \Rightarrow \frac 1 2 gt = v$$$$\Rightarrow t = \frac{2v}{g}$$$$ \Rightarrow x = \frac{2v^2}{g}$$

 

[PeroK]

I would say that means it starts on the first step. If it were to start with a very small nonzero velocity it would land on the second step, etc. But my friend disagrees.

At the top of the steps is the "landing", not the first step!

 

[Henryk]

Your answer 9.1 is (approximately) correct. It means it will fly over 9 steps and land on the 10th.