Question

A stone is dropped from the top of a tower 400 m high and at the same time another stone is projected upward vertically from the ground with velocity of 100 m/sec. Find when and where the two stone will meet?

Answer 1

Answer on Question #50871, Physics, Mechanics Kinematics Dynamics

A stone is dropped from the top of a tower $400\mathrm{m}$ high and at the same time another stone is projected upward vertically from the ground with velocity of $100\mathrm{m / sec}$ . Find when and where the two stone will meet?

Answer:

Firstly we note the given data according to the task. Height of the tower when the stone is dropped from it $H = 400\mathrm{m}$ , velocity of the stone projected upward vertically is equal to 100 m/sec. We need to find when and where the two stone will meet. The graph is shown on the Figure 1.

Figure 1

We consider the downward motion of the first stone:

$$\mathrm {h} = \mathrm {u t} + \frac {1}{2} \mathrm {g t} ^ {2}$$

Let the first stone be dropped from the point A and the second stone be projected from the point C. Suppose the first stone arrives B in time t seconds, where $\mathrm{AB} = (400 - h)m = x$ meters. It is given the second stone also arrives at B after the same time t.

Now for the first stone we have:

$$A B = \frac {1}{2} g t ^ {2} = > x = \frac {1}{2} g t ^ {2}$$

We can express the value of $t$ .

$$t = \sqrt {\frac {2 (4 0 0 - h)}{g}} = \sqrt {\frac {2 (4 0 0 - h)}{9 . 8}}$$

Then we consider the upward motion of the second stone form $C$ to $B$ . Since the stone rises to a height $CB = h$ meters in $t$ seconds. We have the following.

$$\mathrm {h} = \mathrm {u t} - \frac {1}{2} \mathrm {g t} ^ {2}$$ $$\mathrm {h} = (1 0 0 \mathrm {m / s e c}) \mathrm {t} - \frac {1}{2} \mathrm {g} (\sqrt {\frac {2 (4 0 0 - \mathrm {h})}{\mathrm {g}}}) ^ {2}$$

Initial velocity of the second stone $u = 100 \, \text{m/sec}$ , acceleration due to gravity $a = 9.8$

Maximum height is equal to $H_{\max} = \frac{u^2}{2g} = \frac{(100m / sec)^2}{2 \cdot 9.8 \frac{m}{sec^2}} = \frac{10000 \frac{m}{sec^2}}{19.6 \frac{m}{sec^2}} = 510.204 \, \text{m}$

When it is in the middle $S = \frac{H_{\max}}{2} = \frac{510.204}{2} = 255.102 \, \text{m}$

Now we need solve the following equation to find the height $h$ .

$$\mathrm {h} = (1 0 0 \mathrm {m / s e c}) \cdot \sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}} - \frac {1}{2} \mathrm {g} (\sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}}) ^ {2}$$

Simplify the equation.

$$\mathrm {h} = (1 0 0 \mathrm {m / s e c}) \cdot \sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}} - \frac {1}{2} \mathrm {g} \frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}$$ $$\mathrm {h} = (1 0 0 \mathrm {m / s e c}) \cdot \sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}} - (4 0 0 \mathrm {m} - \mathrm {h})$$

We add to both sides of the equation the following expression $(400\mathrm{m} - \mathrm{h})$

$$\mathrm {h} + (4 0 0 \mathrm {m} - \mathrm {h}) = (1 0 0 \mathrm {m / s e c}) \cdot \sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}}$$

Now we can rewrite the equation.

$$(1 0 0 \mathrm {m / s e c}) \cdot \sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}} = 4 0 0 \mathrm {m}$$

Then we can divide both sides of the equation by $\left(\frac{100\mathrm{m}}{\mathrm{sec}}\right)$ .

$$\sqrt {\frac {2 (4 0 0 \mathrm {m} - \mathrm {h})}{\mathrm {g}}} = \frac {4 0 0 \mathrm {m}}{1 0 0 \mathrm {m / s e c}}$$ $$\sqrt{\frac{2(400m - h)}{g}} = 4$$

Then we square both sides of the equation.

$$\frac{2(400 - h)}{g} = 16$$

Now we can find the value of $h$.

$$2(400 - h) = 16g$$ $$400 - h = \frac{16g}{2}$$

Multiply all terms of the equation by -1.

$$h - 400 = -\frac{16g}{2}$$

Then we add $400m$ to both sides of the equation.

$$h = 400 - 8g = 400 - 78.4 = 321.6m$$

Now we can determine the point where two stone will meet.

$$AB = (400 - 321.6)m = 78.4m$$

Then we can find the time. We substitute the determined value of $h$.

$$t = \sqrt{\frac{2(400 - h)}{g}} = \sqrt{\frac{2(400m - 321.6m)}{9.8 \frac{m}{\sec^2}}} = 4 \sec$$

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