Derive an expression for the time taken, (t) for a test mass to fall to the ground from a height (h) in a uniform gravitational field (g = 9.81 ms^-2)

We first build an intuition for exactly what acceleration is and what we expect to happen. In the uniform gravitational field approximation, we assume h to be much less than the radius of the Earth. Therefore, the higher we drop the test mass from, the larger the velocity of the test mass when it hits the ground at h = 0. Derivation: We take velocity = dh/dt = acceleration(g)*time(t), and then use the required calculus to integrate to find the expression: h = (1/2)gt^2 . We finally tidy this up by rearranging using basic algebra to express t as a function of h t(h) = SQRT(2h/g) (I would then very likely ask to plot that function so that the intuitions developed at the start can be solidified and we can gain some confidence in the use of calculus to solve real problems.)