A life form standing on the surface of an unknown planet throws a small inanimate sphere vertically upwards and then steps backwards. The life form releases the sphere at a height of 1m; after 4 seconds the sphere has reached a height of 37 m, after a further 4 seconds the sphere is at a height of 25 m.
Derive a quadratic function, h(t), relating the height of the sphere to the elapsed time.
The height of the ball $h$ is a quadratic function of time $t$ meaning $$h=a\,t^2 + b\,t+c$$ To get the coefficients $a$,$b$ and $c$ we plug in the known values of $h(t)$, i.e. initially the ball is at 1m which means $h(0)=1$ therefore $h(0)=c=1 \Rightarrow c=1$. Now we use the next point to calculate b (in terms of a) $$h(4)=37 = 16a+4b \Rightarrow b = \frac{37-16a}{4}$$ The final point gives us $$h(8)=25=625a+25b \Rightarrow 1=25a+b$$ into which we can substitute $b$ from above giving $a = -\frac{11}{28}$ and then substituting this into the equation for $b$ gives $b=\frac{303}{28}$ thus
$$ h(t)=-\frac{11}{28}t^2 + \frac{303}{28}t+1$$
To see that this makes sense note that the negative quadratic term indicates in inverted parabola as expected and the positive linear term corresponds to the ball having a positive initial velocity.