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Need physics help 10 points

[From: ] [author: ] [Date: 12-09-23] [Hit: ]
If the water splash is heard 3 seconds after the rock was dropped, and that the speed of sound is 1100 ft / sec, approximate the height of the well. T1 = 2. H = 16 T1 2 = 16 (2.88) 2 = 132.......
A rock is dropped into a well and the distance travelled is 16 t^2 feet, where t is the time. If the water splash is heard 3 seconds after the rock was dropped, and that the speed of sound is 1100 ft / sec, approximate the height of the well.

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T1 - time rock take to reach t bottom of th well
H is height of the well

H = 16 T1 2

T2 - time sound takes wave to reach the top of th well

H = 1100 T2

relationship between T1 and T2 is

T1 + T2 = 3

eliminate H & combine th equations H = 16 T1 2 & H = 1100 T2 to get

16 T1 2 = 1100 T2

to substitute T2 by 3 - T1 in above equation

16 T1 2 = 1100 (3 - T1)

above is a quadratic equation tht can be written as

16 T1 2 + 1100 T1 - 3300 = 0

above equations has 2 solutions and only 1 is positive and is givn by

T1 = 2.88 seconds

now to calculate height of the well

H = 16 T1 2 = 16 (2.88) 2 = 132.7 feet
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