Now that we have learned the basics of both horizontal and vertical motion, it is
time to put them together. There is really nothing new to be learned here, we just need
to be sure that we don't cross variables. The only thing that can be used in both sets
of equations, since it is the same for both, is time.
A young prankster wishes to hit a truck traveling at a constant speed of 20m/s with
a water balloon. If the prankster is in a tree, 40m from the top of the truck, how far
should the truck be from the tree when she releases the balloon so that it hits the
truck when it is at a position directly beneath her?
First, we need to find out how long it will take the balloon to fall.
-40m = 0m/s*t + 0.5*-9.8m/s2*t2
-40m = -4.9m/s2*t2
8.163s2 = t2
2.857s = t
Now that we know how long it will take for the balloon to drop, we need to decide
how far out the truck could be. Since we know how fast it is going, that will be easy.
20m/s * 2.857s = d
57.140m = d
The truck needs to be 57.14m from the prankster when she lets the balloon drop.
This lesson dealt with with horizontal and vertical motion working together, each on
separtate objects. In the next few lessons we will be
dealing with projectile motion, which occurs when horizontal and vertical forces act
upon the same object at the same time.
Vertical Motion Sample Problems
