Question

If a stone is thrown up at 10 m per second from a height of 100 meters above the surface of the moon, its height in meters after t seconds is given by s = 100+10t-0.8t^2 . What is its acceleration?

Answer 1

-1.6 m/s

Explanation:

According to the equations of motion, the height in meters (s) after t seconds is:

`s(t) = s_0+v_0t+a(t^2)/(2)`

If the initial position is 10 m and the initial velocity is 10 m/s:

`s(t) = 100+10+a(t^2)/(2)`

When comparing it with the given height equation, it is possible to obtain the acceleration as follows:

`100+10+a(t^2)/(2) =100+10t-0.8t^2\\(a)/(2)=-0.8\\ a=-1.6\ m/s`

Acceleration is -1.6 m/s.