Posted by Austin on .
how to find the max height of a cat that is thrown at 9.8 m/s and is in the air for a total of 4 seconds.

Physics 
Steve,
If thrown from height h=0,
h(t) = 9.8t  4.9t^2
this is a parabola with vertex where t=1.
So, plug that in to get h(1)
Now, you say the cat flew for 4 seconds. So, the cat must have been thrown from some height H. So,
h(t) = H + 9.8t  4.9t^2
has a root at t=4. So,
h(4) = 0 means
9.8(4)  4.9(16) + H = 0
H = 39.2
So, add 39.2 to h(1) as indicated above.