Question

A car driver travelling at 90km/h, notices that he is heading directly to a cliff when he is 20m away from the edge. He applies the brakes and experiences a backward acceleration of 8.0 m/s2, but it's not enough. The car goes off the cliff, if the cliff is 12m high, how far is the car traveling horizontally until it lands?

Answer 1

Question #65279, Physics / Mechanics | Relativity

A car driver travelling at $90\mathrm{km/h}$, notices that he is heading directly to a cliff when he is 20m away from the edge. He applies the brakes and experiences a backward acceleration of $8.0~\mathrm{m/s^2}$, but it's not enough. The car goes off the cliff, if the cliff is 12m high, how far is the car traveling horizontally until it lands?

$$\begin{array}{l} v_0 = 90 \frac{\mathrm{km}}{\mathrm{h}} = 25 \frac{\mathrm{m}}{\mathrm{s}} \\ s = 20 \mathrm{m} \\ a = 8 \mathrm{m/s^2} \\ h = 12 \mathrm{m} \\ x = ? \\ \end{array}$$

The answer to the question.

Equations of uniformly accelerated motion:

$$\left\{ \begin{array}{l} v = v_0 - a t \\ s = v_0 t - \frac{a t^2}{2}; \end{array} \right.$$

We find the movement time:

$$\begin{array}{l} \frac{a t^2}{2} - v_0 t + s = 0 \\ \frac{8 t^2}{2} - 25 t + 20 = 0 \\ 4 t^2 - 25 t + 20 = 0 \\ t = \frac{25 - 17.5}{8} = 0.9 \mathrm{s}; \end{array}$$

The second root of the equation does not satisfy the condition of the problem.

$$v = 25 - 7.5 = 17.5 \frac{\mathrm{m}}{\mathrm{s}};$$

Consider the uniformly accelerated drop $h$.

$$h = \frac{g t_1^2}{2}; \quad \text{hence} \quad t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{24 \mathrm{m}}{9.8 \mathrm{m/s^2}}} = 1.6 \mathrm{s}$$ $$x = v \quad t = 17.5 \frac{\mathrm{m}}{\mathrm{s}} \cdot 1.6 \mathrm{s} = 27.4 \mathrm{m}$$

Answer: the car traveling horizontally until it lands $x = 27.4$ m.

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