Question

A falling object requires 1.50s to travel the last 30m before hitting the ground. From what height above the ground did it fall?

Answer 1

Answer on Question #73874 Physics / Mechanics | Relativity

A falling object requires $\tau = 1.50$ s to travel the last $s = 30$ m before hitting the ground. From what height above the ground $h$ did it fall?

Solution:

Let us denote as $t$ the total time of object motion. Then

$$h = \frac{g t^2}{2}$$

and

$$h - s = \frac{g (t - \tau)^2}{2}$$

These equations give

$$s = \frac{g t^2}{2} - \frac{g (t - \tau)^2}{2} = g \tau t - \frac{g \tau^2}{2}$$

So

$$t = \frac{s + \frac{g \tau^2}{2}}{g \tau} = \frac{s}{g \tau} + \frac{\tau}{2} = \frac{30}{9.8 \times 1.50} + \frac{1.50}{2} = 2.79 \text{ s}$$

Thus, the height

$$h = \frac{g t^2}{2} = \frac{9.8 \times 2.79^2}{2} = 38.16 \text{ m}$$

Answer: 38.16 m

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