QuestionA soccer player kicks a rock horizontally off a $35.0-\mathrm{m}$ -high cliff into a pool of water. If the player hears the sound of the splash 2.86 s later, what was the initial speed given to the rock? Assume the speed of sound in air is $343 \mathrm{~m} / \mathrm{s}$ . $\square$ The distance traveled by the sound is the hypotenuse of the triangle formed by the height and the horizontal distance. $\mathrm{m} / \mathrm{s}$

Studdy Solution

STEP 1

1. The height of the cliff is $35.0 \, \mathrm{m}$ .
2. The time taken to hear the splash is $2.86 \, \mathrm{s}$ .
3. The speed of sound in air is $343 \, \mathrm{m/s}$ .
4. The rock is kicked horizontally, meaning the initial vertical velocity is zero.
5. The horizontal distance traveled by the rock is unknown and needs to be determined.
6. The time the sound takes to travel from the water back to the player is a contributing factor to the total time.

STEP 2

1. Calculate the time it takes for the rock to fall the vertical distance of $35.0 \, \mathrm{m}$ .
2. Determine the time it takes for the sound to travel back to the player.
3. Use the total time and the time for the sound to find the time the rock was in free fall.
4. Calculate the horizontal distance traveled by the rock.
5. Determine the initial horizontal speed given to the rock.

STEP 3

Use the equation of motion for free fall to calculate the time it takes for the rock to fall $35.0 \, \mathrm{m}$ . The equation is: $h = \frac{1}{2} g t^2$ where $h = 35.0 \, \mathrm{m}$ and $g = 9.81 \, \mathrm{m/s^2}$ .
Solve for $t$ : $35.0 = \frac{1}{2} \cdot 9.81 \cdot t^2$ $t^2 = \frac{35.0 \cdot 2}{9.81}$ $t^2 = 7.134 \, \mathrm{s^2}$ $t = \sqrt{7.134}$ $t \approx 2.67 \, \mathrm{s}$

STEP 4

Subtract the time of free fall from the total time to find the time it takes for the sound to travel back to the player: $t_{\text{sound}} = 2.86 \, \mathrm{s} - 2.67 \, \mathrm{s}$ $t_{\text{sound}} \approx 0.19 \, \mathrm{s}$

STEP 5

Calculate the horizontal distance the sound travels using the speed of sound: $d = v_{\text{sound}} \cdot t_{\text{sound}}$ $d = 343 \, \mathrm{m/s} \cdot 0.19 \, \mathrm{s}$ $d \approx 65.17 \, \mathrm{m}$

STEP 6

Now that we have the horizontal distance $d$ , calculate the initial horizontal speed $v_0$ given to the rock using the time of free fall: $v_0 = \frac{d}{t}$ $v_0 = \frac{65.17 \, \mathrm{m}}{2.67 \, \mathrm{s}}$ $v_0 \approx 24.4 \, \mathrm{m/s}$
The initial speed given to the rock is approximately $24.4 \, \mathrm{m/s}$ .

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