Which step does the car hit when it lands?

Question

Little Jin rolls his toy car horizontally off the top of a staircase with a speed of 3.0m/s. Each of the steps is 0.18 m high and 0.30 m wide. Which step does the car hit when it lands?

Answer 1

Let Vo=initial velocity, h=drop height and S=horizontal distance. Then,
S=Vo*t (Vo=3 m/s) ----(1)
h=0.5*g*t^2 (g= 10 m/s^2)
t=sqrt(2*h/g) ---(2)
Insert (2) in (1) and solve for S
S=Vo*sqrt(2*h/g)----(3)
On the other hand, h and S are in 3:5 proportion, i.e.,
S=5/3*h---(4)
Equating (3) and (4) and solving for h, we get
h= 0.65 m,a nd
S=1.08 m
to get # of steps, N
N=h/0.18=S/0.3
N=3.6
Therefore the toy car hits the forth step.

Answer 2

Lol, I have a physics test tomorrow so I better be able to do this.


Ok, the ratio of stair height and width is .18/.30

This is also the ratio of Y position over X position.

Now, X(t) = x0 + vt + 1/2at^2; x0= 0, 1/2at^2 = 0 because it starts at origin and has no acceleration. vt = 3

Y(t) = y0 + vt+ -1/2gt^2; y=0 vt = 0 but g = 9.8

ratio = Y(t)/X(t) = 4.9t^2/3t; Solve for T. That is the time when it hits a staircase. T = .3673

now, after you find T, you know that 3t= x position displaced by particle movement. You also know the width of each step.

plug t into 3t and then divide by width of the steps. Round Up to whatever whole number is closest and that should equal 4.

3.67 meters from origin..round up to 4.

Someone else had the exact same number values for the qeustion...strange lol

Answer 3

Hard to explain, but you'll wind up with a table that looks like:

Step 1 0.18m drop 0.1916s time 0.5748m travelled 0.3m step distance

Step 2 0.36m drop 0.2710s time 0.8129m travelled 0.6m step distance

Step 3 0.54m drop 0.3319s time 0.9956m travelled 0.9m step distance


Calculate the total drop with each additional step. Determine the time required to fall that distance. Multiply fall time with initial velocity determine if the distance travelled is longer than the width of the steps travelled. You can see the step 3 is almost there