The Suitcase Chute

## Question:

A 40 lb suitcase slides from rest down a 20 foot smooth ramp with
a incline of 30 degrees. The foot of the ramp is 4 feet above the
floor. At what horizontal distance from the foot of the ramp does
the suitcase hit the floor and how long does the total trip take?
## Answer:

Break the problem into two parts, what happens on the slide and
what happens after it leaves the slide.
The acceleration(a) of the suitcase along the slide is the
acceleration due to gravity times the sine of the incline angle,
or a=32.2*sin(30)=16.1f/s/s. Starting
with zero velocity, the suitcase distance(d) from the top of the
slide is

d=1/2*a*t^{2}.

At d=20ft,

40=16.1*t^{2}, or t^{2}=2.48s^{2}, or t=1.58s.

Neglecting the dimensions of the suitcase so we do not have to
deal with the rotation that occurs when the leading edge of the
suitcase clears the slide by 1/2 its length, the velocity at the
instant the suitcase leaves the slide is
16.1f/s/s*1.58s=25.44 f/s directed 30 degrees below
horizontal. The vertical component of that velocity is

Vv=25.44*sin(30)=12.72f/s.

The horizontal component is

Vh=25.44*cos(30)=22.03f/s.

The vertical distance(Dv) fallen in time, t, after leaving the
slide is

Dv=12.72*t+16.1*t^{2}=4.

The time to hit the floor then is given by

4=12.72*t+16.1*t^{2}, or 16.1*t^{2}+12.72*t-4=0.

Solving the quadratic equation,

t=(-12.72(+/-)(12.72^{2}-4*16.1*(-4))^{.5})/32.2,
or

t=(-12.72(+/-)20.48)/32.2, or t=0.2425,

taking the positive value of t.

The total time to go from the top of the slide to the floor is
1.58+0.2425=1.82s.

The horizontal distance covered after leaving the slide will
be the horizontal velocity times the fall time or

Dh=22.03f/s*0.2425s=5.34ft.

This information is brought to you by M. Casco Associates, a
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understanding how the universe works. My name is James D. Jones.
If I can be of more help, please let me know. JDJ