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202301161044 Exercise 3.5.12

A stone is thrown upwards with a velocity of u\,\mathrm{m\,s^{-1}} at the same instant that a stone is let fall from the top of a cliff immediately above the first stone. They meet after t\,\mathrm{s}. What is the height of the cliff?

Extracted from L. H. Clarke & F. G. J. Norton. (1972). Additional Applied Mathematics.

Roughwork.

Let the height of the cliff be H\,\mathrm{m}. Take upward positive, and the ground s=0 the reference level.

For the stone let fall, parametrize its trajectory:

\mathbf{s}\Big(t\in \Big[0,\sqrt{\frac{2H}{g}}\Big]\Big)=\displaystyle{\bigg(H-\frac{1}{2}gt^2\bigg)\,\hat{\mathbf{j}}},

and for the stone thrown upward, also:

\mathbf{s}\big(t\in \big[ 0,\frac{2u}{g}\big]\big)=\displaystyle{\bigg(ut-\frac{1}{2}gt^2\bigg)\,\hat{\mathbf{j}}}.

As the two stones meet at time t=t':

\begin{aligned} H-\frac{1}{2}gt'^2 & = ut'-\frac{1}{2}gt'^2 \\ H & = ut' \end{aligned}

This problem is not to be attempted.