Vertical motion problem?

Question

A cat is thrown vertically upward and hits the ground 15 seconds later.

A) How hight does the cat travel upward?

B) What is its velocity when it strikes the ground?

Answer 1

Assuming that the cat began its trip on the ground:

Exactly half the time is spent going up, and the other half going down, so the downward part of its trip takes 7.5 seconds.

The position equation is:

h(t) = h(0) + v(0)*t + 1/2*a*t^2

Since we are only considering the downward portion of the cat's trajectory, v(0) is 0, and to simplify our calculations, let's consider the ground as having 0 height, so that's the final value for our equation.

We now have:

0 = h(0) + 1/2*a*t^2

Since we are on Earth, a = -9.8 m/s^2, and the cat traveled for 7.5 seconds.

0 = h(0) - 4.9*(7.5^2)

I'm sure that you can run the numbers and solve for h(0) from here.

As for b, the velocity equation is:

v(t) = v(0) + a*t

Again, since we are only considering the downward portion of the flight, v(0) is 0 and t is 7.5s.  a is still -9.8.  All you need to do is plug in the numbers.