I haven't done this in some time, and I was horrible with mechanic, but here you go.

This is a pretty straightforward problem that can be solved with classical mechanics.

We'll use this equation.

Y=Yi + Vy * T -1/2 * a * T^2

where

Y is the height of the box after T seconds

Yi is the initial height before you throw the box (If you throw the box from 98 feet high, and the car is 4 feet high, it has to travel 94 feet to hit it, so Yi=94 feet )

Vy is the initial vertical speed of the box (0 in that case

T is the time in seconds

a is the acceleration of the box (the acceleration given by the force of gravity : 9.8 m/s^2 or 32 feet/s^2)

When the box will hit the car, Y will be equal to 0.

0=94 feet + 0 feet *T sec - 1/2 * 32 feet * T^2 square sec

You have only one unknown value therefore you can solve it.

-94 =-16T^2

5.975 = T^2

so T is equal to either 2.4 or -2.4 seconds. You can reject the -2.4 second since it's impossible for the box to hit the car before you threw it :).

And if you want to know whose car you will hit, you can use this equation.

X = Vx * T

Where

X is the distance the box will travel in T seconds

Vx is the initial speed of the box (25 feet / seconds)

T is the time (the 2.4 seconds we found in the other equation)

X sec = 25 feet/sec * 2.4 sec

X = 60 feet

So the owner of the car that is 60 horizontal feet away from your room will be the new owner of your broken cable box!

Anyway I didn't factor the air resistance because I don't know it by heart (and I didn't feel like looking for it).