Question

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.63 m. The stones are thrown with the same speed of 8.97 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

Answer 1

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is $6.63\mathrm{m}$ . The stones are thrown with the same speed of $8.97\mathrm{m/s}$ . Find the location (above the base of the cliff) of the point where the stones cross paths.

Solution:

$V_{1} = 8.97\frac{\mathrm{m}}{\mathrm{s}} -$ velocity of the stone, which was thrown up;

$V_{2} = 8.97\frac{\mathrm{m}}{\mathrm{s}} -$ velocity of the stone, which was thrown down;

$\mathrm{H} = 6.63\mathrm{m}$ - height of the cliff;

h - height of the point where the stones cross paths (above the base of the cliff).

The equation of motion for the first stone (which was thrown up) respect to the Y-axis:

$$y _ {1} = V _ {1} t - \frac {g t ^ {2}}{2}$$

The equation of motion for the second stone (which was thrown down) respect to the Y-axis:

$$y _ {2} = H - V _ {2} t - \frac {g t ^ {2}}{2}$$

When stones the stones cross paths, their coordinates are equal:

$$y _ {2} = y _ {1}$$

(3) and (2) in (1):

$$V _ {1} t _ {\text {c r o s s}} - \frac {g t _ {\text {c r o s s}} ^ {2}}{2} = H - V _ {2} t _ {\text {c r o s s}} - \frac {g t _ {\text {c r o s s}} ^ {2}}{2}$$ $$\mathrm {H} = \mathrm {t} _ {\text {c r o s s}} \left(\mathrm {V} _ {2} + \mathrm {V} _ {1}\right)$$ $$t _ {\text {c r o s s}} = \frac {H}{V _ {2} + V _ {1}}$$ $$\begin{array}{l} h = y _ {2} \left(t _ {\text {c r o s s}}\right) = y _ {1} \left(t _ {\text {c r o s s}}\right) = V _ {1} t _ {\text {c r o s s}} - \frac {g t _ {\text {c r o s s}} ^ {2}}{2} = \frac {V _ {1} H}{V _ {2} + V _ {1}} - \frac {g H ^ {2}}{2 \left(V _ {2} + V _ {1}\right) ^ {2}} = \\ = \frac {H}{V _ {2} + V _ {1}} \left(V _ {1} - \frac {g H}{2 \left(V _ {2} + V _ {1}\right)}\right) = \frac {6 . 6 3 m}{8 . 9 7 \frac {m}{s} + 8 . 9 7 \frac {m}{s}} \left(8. 9 7 \frac {m}{s} - \frac {9 . 8 \frac {m}{s ^ {2}} \cdot 6 . 6 3 m}{2 \left(8 . 9 7 \frac {m}{s} + 8 . 9 7 \frac {m}{s}\right)}\right) \\ = 2. 6 5 m \\ \end{array}$$

This means that paths will cross on a hight $h = 2.65m$ where each Y-coordinate is $2.65m$ ( $y_2 = y_1 = 2.65m$ ).

The location (above the base of the cliff) of the point where the paths of the stones will cross:

$$\mathrm{h} = 2.65 \, \mathrm{m}$$

Answer: paths will cross at the high $\mathrm{h} = 2.65 \, \mathrm{m}$ above the base of the cliff.