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

Answer:

-1.6 m/s

Step-by-step explanation:

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

$s(t) = s_0+v_0t+a\frac{t^2}{2}$

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

$s(t) = 100+10+a\frac{t^2}{2}$

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

$100+10+a\frac{t^2}{2} =100+10t-0.8t^2\\\frac{a}{2}=-0.8\\ a=-1.6\ m/s$

Acceleration is -1.6 m/s.