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?

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*t2.
At d=20ft,
40=16.1*t2, or t2=2.48s2, 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*t2=4.
The time to hit the floor then is given by
4=12.72*t+16.1*t2, or 16.1*t2+12.72*t-4=0.