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A lead ball is dropped into a lake from a diving board 5.0 meters above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 seconds after it is released. How deep is the lake?

Respuesta :

pablomai290696

Answer:

28.8 meters

Explanation:

We must first determine at which velocity the ball hits the water. To do so we will:

1) Assume no air resistance.

2) Use the Law of conservation of mechanical energy: E=K+P

Where

E is the mechanical energy (which is constant)

K is the kinetic energy.

P is the potential energy.

With this we have [tex]\frac{m}{2} *v^{2}  = m*g*h[/tex]

Where:

m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.

v is the speed when it hits the water.

g is the gravitational constant (we will assume g=9.8[tex]\frac{m}{s^{2} }[/tex].

h is the height from which the ball fell.

Because when we initially drop the ball, all its energy is potential (and [tex]P = - m*g*h[/tex]) and when it hits the water, all its energy is kinetic ([tex]K=\frac{m}{2} *v^{2}[/tex]. And all that potential was converted to kinetic energy.

Now, from [tex]\frac{m}{2} *v^{2}  = m*g*h[/tex] we can deduce that [tex]v=\sqrt{2*g*h}[/tex]

Therefore v=9.6[tex]\frac{m}{s}[/tex]

Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.

So D=9.6[tex]\frac{m}{s}[/tex]*3[tex]s[/tex]=28.8[tex]m[/tex]

Which is our answer.

snehashish65

The depth of lake is 29.67 m.

Given data:

The distance of diving board above the water is, d = 5.0 m.

The time interval to reach the bottom is, t = 3.0 s.

We need to obtain the velocity of ball after hitting the water. This uses the conservation of Energy, which says that at any point, the kinetic energy and the potential energy remains constant. Then using the conservation of Energy as,

[tex]PE = KE\\\\mgd = \dfrac{1}{2}mv^{2}\\\\gd = \dfrac{1}{2}v^{2}\\\\v =\sqrt{2 \times g\times d}\\\\v =\sqrt{2 \times 9.8\times 5.0}\\\\v = 9.89 \;\rm m/s[/tex]

Now, the depth of the lake is calculated as,

[tex]d' = v \times t[/tex]

here, d' is the depth of the lake.

Solving as,

[tex]d' = 9.89 \times 3\\\\d' = 29.67 \;\rm m[/tex]

Thus, we can conclude that the depth of lake is 29.67 m.

Learn more about the conservation of energy here:

https://brainly.com/question/2737162

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