Posted by **Austin** on Thursday, October 10, 2013 at 5:40pm.

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**, Thursday, October 10, 2013 at 5:57pm
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.

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