Question

17 An object is thrown upward from the edge of a building with a velocity of 20 m/s. Where will the object be 3s after it was thrown?


−22m
16m

22m


−16m

Answer 1

Answer on Question#52032 - Physics - Quantum Mechanics

17. An object is thrown upward from the edge of a building with a velocity of $20 \, \text{m/s}$. Where will the object be 3s after it was thrown?

- -22m

- 16m

- 22m

- -16m

**Solution:**

The dependence of the height on time is given by

$$h(t) = v_0 t - \frac{g t^2}{2},$$

where $v_0$ – is the initial velocity, $g = 9.8 \, \frac{\text{m}}{\text{s}^2}$ – is the acceleration of free fall.

Since $v_0 = 20 \, \frac{\text{m}}{\text{s}}$, we obtain

$$h(3s) = 20 \, \frac{\text{m}}{\text{s}} \cdot 3s - \frac{9.8 \, \frac{\text{m}}{\text{s}^2} \cdot 9\text{s}^2}{2} = 16\text{m}$$

**Answer:** 16m.