QuestionA bowling ball traveling with constants speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is $340 \mathrm{~m} / \mathrm{s}$ ?

Studdy Solution

STEP 1

1. The length of the bowling lane is 16.5 meters.
2. The total time taken to hear the sound of the ball hitting the pins after release is 2.80 seconds.
3. The speed of sound is 340 meters per second.
4. The ball travels at a constant speed from the release point to the pins.

STEP 2

1. Define the variables and equations related to the problem.
2. Calculate the time it takes for the sound to travel back to the bowler after hitting the pins.
3. Determine the time it takes for the ball to reach the pins.
4. Calculate the speed of the ball using the distance and time obtained.

STEP 3

Define the variables and set up the equations.
Let $v_b$ be the speed of the ball. The total distance the ball travels is $d = 16.5 \, \text{m}$ . The speed of sound is $v_s = 340 \, \text{m/s}$ . The total time from release to hearing the sound is $t_{total} = 2.80 \, \text{s}$ . Let $t_b$ be the time taken by the ball to reach the pins. Let $t_s$ be the time taken by the sound to travel back to the bowler.

STEP 4

Calculate the time it takes for the sound to travel back to the bowler after hitting the pins.
The sound travels the same distance $d = 16.5 \, \text{m}$ .
$t_s = \frac{d}{v_s}$
$t_s = \frac{16.5 \, \text{m}}{340 \, \text{m/s}}$
$t_s \approx 0.0485 \, \text{s}$

STEP 5

Determine the time it takes for the ball to reach the pins.
The total time $t_{total}$ is the sum of the time taken by the ball to reach the pins and the time taken by the sound to travel back.
$t_{total} = t_b + t_s$
$t_b = t_{total} - t_s$
$t_b = 2.80 \, \text{s} - 0.0485 \, \text{s}$
$t_b \approx 2.7515 \, \text{s}$

STEP 6

Calculate the speed of the ball using the distance and time obtained.
$v_b = \frac{d}{t_b}$
$v_b = \frac{16.5 \, \text{m}}{2.7515 \, \text{s}}$
$v_b \approx 5.998 \, \text{m/s}$
Rounding to appropriate significant figures, we obtain:
$v_b \approx 6.00 \, \text{m/s}$
The speed of the ball is approximately $6.00 \, \text{m/s}$ .

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