A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air is 330 m/s.

What was the average speed of the base - ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.)

Question

Physics

Armani Wang

23 October, 23:15

A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air is 330 m/s.

What was the average speed of the base - ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.)

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Answers (1)

Know the Answer?

Manuel Marks · Answer 1

Speed = 52/1.57576 = 33 m/s

Explanation:

Distance = speed * time

Given, the distance traveled by the baseball = 52 m

Speed of sound in air = 330 m/s

Total time = 1.73333 s.

Total time for the student to hear the sound of the gong is the sum of the time take for the baseball to reach the gong and the time taken by the sound to travel back.

Distance traveled by the sound is 52 m and the speed is 330 m/s

So time taken by the sound to travel back = distance traveled / the speed

=> time = 52/330 s = 0.15757 s

Time taken by the base ball to reach the gong is the total time - the time taken by the sound

=> time taken by base ball = 1.73333 - 0.15757 = 1.57576s

Speed of the base ball to reach the gong = distance / time

Speed = 52/1.57576 = 33 m/s

A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air is 330 m/s.What was the average speed of the base - ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.)

A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air is 330 m/s.

What was the average speed of the base - ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.)