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
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