SOLUTION: On the moon, the acceleration due to gravity is constant 1.6 {{{ m/s^2 }}}, i.e. a(t) = -1.6. If a ball is thrown into the air with an initial velocity of 2{{{ m/s }}} at an init

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Question 1031099: On the moon, the acceleration due to gravity is constant 1.6 +m%2Fs%5E2+, i.e. a(t) = -1.6. If a ball is thrown into the air with an initial velocity of 2+m%2Fs+ at an initial height of 1 meter, find the height as a function of time, h(t).
All I know is that derivative of velocity is acceleration and the anti-derivative is the position. I don't know how to tie the elements together though.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
On the moon, the acceleration due to gravity is constant 1.6 +m%2Fs%5E2+, i.e. a(t) = -1.6. If a ball is thrown into the air with an initial velocity of 2+m%2Fs+ at an initial height of 1 meter, find the height as a function of time, h(t).
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h(t) = at^2/2 + vo*t + ho --- vo = start speed, ho = start height
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h(t) = -0.8t^2 + 2t + 1

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
As I know, there is no air on the moon.