Question

A worker drops a wrench from the top of a tower 87.0 m tall. What is the velocity when the wrench strikes the ground?

Answer 1

1. A worker drops a wrench from the top of a tower $87.0\,\mathrm{m}$ tall. What is the velocity when the wrench strikes the ground?

And the velocity depends linearly on time:

$$v = a t.$$

Lets express the time of the falling from the Eq. (1) and substitute it into the Eq. (2).

$$h = \frac{g\,t^2}{2}, \quad t = \sqrt{\frac{2h}{g}}.$$ $$v = g \cdot \sqrt{\frac{2h}{g}} = \sqrt{2\,g\,h}.$$

So, the velocity of the wrench, when it strikes the ground, is: $\boxed{v = \sqrt{2\,g\,h}}$.

Let check the dimension.

$$[v] = \sqrt{\frac{\mathrm{m}}{\mathrm{s}^2} \cdot \mathrm{m}} = \sqrt{\frac{\mathrm{m}^2}{\mathrm{s}^2}} = \frac{\mathrm{m}}{\mathrm{s}}.$$

Let evaluate the quantity.

$$v = \sqrt{2 \cdot 9.8 \cdot 87} \approx 41.3\left(\frac{\mathrm{m}}{\mathrm{s}}\right).$$

Answer: $41.3\,\frac{\mathrm{m}}{\mathrm{s}}$.