Question

One ball is thrown upwards with an initial velocity of 20 m/s from the edge of a tall building. Three seconds later, a ball is dropped from the edge of the roof. Both balls hit the ground at the same time. How tall is the building?

Answer 1

Question #54576, Physics / Mechanics | Kinematics | Dynamics

One ball is thrown upwards with an initial velocity of $20\,\mathrm{m/s}$ from the edge of a tall building. Three seconds later, a ball is dropped from the edge of the roof. Both balls hit the ground at the same time. How tall is the building?

Answer.

Let H be the high of the building.

1-st ball

Time to highest point: $v_0 - gt = 0 \rightarrow t = \frac{v_0}{g}$

high of this point from the roof: $h = \frac{gt^2}{2} = \frac{v_0^2}{2g}$

time from this point to the ground:

$$h + H = \frac{gt^2}{2} \rightarrow t = \sqrt{\frac{2(H + h)}{g}} = \sqrt{\frac{2H}{g} + \frac{v_0^2}{g^2}}$$

2-nd ball

time from roof to the ground:

$$H = \frac{gt^2}{2} \rightarrow t = \sqrt{\frac{2H}{g}}$$

Thus $\sqrt{\frac{2H}{g} + \frac{v_0^2}{g^2}} + \frac{v_0}{g} = \sqrt{\frac{2H}{g}} + 3$

For $v_0 = 20\,\frac{m}{s}$, $g = 9.81\,\frac{m}{s^2}$ we have

$$\begin{array}{l} \sqrt{0.204H + 4.156} = \sqrt{0.204H} + 0.961 \rightarrow \\ \rightarrow 0.204H + 4.156 = 0.204H + 1.992\sqrt{0.204H} + 0.924 \rightarrow \\ \rightarrow 1.662 = \sqrt{0.204H} \rightarrow H = 13.54\,\mathrm{m}. \end{array}$$

The building is 13.54 m high.

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