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?
Expert's answer

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)



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



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


Simplifying the expression, we obtain the quadratic equation

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

Solve the quadratic equation

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



Hence t'=1.114s. Find answers to questions



2) The stone is thrown up.



Answer. 1)


2) The stone is thrown up.



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