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)

$$h_0=5.39m$$