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

Question

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?

Answer 1

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

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

Solve the quadratic equation

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

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

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

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)

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

Answer. 1)

$v_0=10.91m/s$

2) The stone is thrown up.

3)

$h_0=5.39m$