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The assignment is to form a function $h(t)$ where $h$ is height of liquid inside a container and $t$ is time.

The volume of the liquid is $V=Ah$ where A is a known constant - radius of the bottom of the container and $h$ is the current height of the liquid.

$h_0$ is a known constant, $h(0)=h_0$. The liquid is draining away from the container at the speed $q$, where $q=kh$. $q_0$ is the drain speed at $h_0$ and $t=0$, and k is a constant multiplier that can be calculated using $q_0$ and $h_0$.

So in other words, the liquid is draining at the pace of $q$, where $q$ is based on $h$.

With these criteria I am to form the function $h(t)$, at first degree, so that I can determine the height of the liquid at any given time.

My own function ended at being $h(t)=h_0-khA^{-1}$, but the problem is that $khA^{-1}$ has $h$ in it so I cannot compute the function without solving the equation first, which is not desirable.

My question is that does my initial function seem right in the first place, and how could I eliminate the h variable from the function and get time variable into it.

The known constants: $h_0, A, q_0, k $

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    It looks like you should be setting up a differential equation. The rate of change of volume which is proportional to the rate in change of the height of the liquid is proportional to the speed q at which you know depends on h. So you have $\frac{dh(t)}{dt} = const * h(t)$ – 2017-02-16
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    Does this mean something along the lines of $q=\frac{dh(t)}{dt}=kh(t)$ ? – 2017-02-16
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    Should read $A\frac{dh}{dt}$ as it is the rate of change of volume. A corresponds to area of bottom of the container, not radius as you write. – 2017-02-16

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I was able to get the answer elsewhere, but here is basically how it was spun to hopefully the correct form:

$$d/dt V(t) = A d/dt h(t) = q(t)$$ $$A d/dt h(t) = kh(t)$$ $$h(t) = h_0e^{-\frac{t}{5}}$$