# Answer to Question #103742 in Mechanics | Relativity for Rethabile Force

Question #103742
From a promontory overhanging a lake, you throw a stone that enters the water vertically at 15.0 m/s You hear the splash 2.66 s
after you release the stone. The speed of sound in air is 340 m/s.

1.What was the stone's initial speed?

2.Did you throw the stone down or up?

3.From what height above the lake surface did you release the stone?
1
2020-02-26T10:12:36-0500

Solution. If a body freely falls from a height, its velocity and height can be described by the formulas (g=9.8 m/s^2 is acceleration of gravity)

"v=gt"

"h=\\frac{gt^2}{2}"

According to the condition of the problem, the velocity of the body during impact 15.0 m/s. Therefore get time and height

"t_1=\\frac{v}{g}=\\frac{15}{9.8}\\approx 1.53s"

"h_1=\\frac{9.8 \\times 1.53^2}{2}\\approx11.47m"

Hence, time for sound

"t_2=\\frac{11.45}{340}\\approx 0.0337s""t_1+t_2<2.66"

As result get that the body moves with a initial speed of v0 from from initial height of h0 to a height h' and then freely falls. Let t' is body lift time. Therefore velocity and height can be represented as

"v_0=gt'"

"h'=\\frac{gt'^2}{2}"

The total time of motion of the body can be represented by the equation

"1.53+t'+\\frac{11.47-h'}{340}=2.66"

Simplifying the expression, we obtain the quadratic equation

"0.0144 t'^2-t'+1.096=0"

"D=(-1)^2-4\\times 1.096\\times0.0144=0.93687=0.9679^2"

"t'_1=\\frac{1-0.9679}{2\\times0.0144}\\approx1.114s"

"t'_2=\\frac{1+0.9679}{2\\times0.0144}\\approx68.33>2.66"

Hence t'=1.114s. Find answers to questions

1)

"v_0=gt'=9.8\\times1.114s=10.91m\/s"

2) The stone is thrown up.

3)

"h_0=11.47-\\frac{gt'^2}{2}\\approx5.39m"

"v_0=10.91m\/s"

2) The stone is thrown up.

3)

"h_0=5.39m"

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