Answer:
[tex]K.E=29.403125J[/tex]
Explanation:
From the question we are told that
Mass [tex]M=100[/tex]
Height [tex]50-20=30m[/tex]
Generally the equation for velocity before impact is is is mathematically given by
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2*9.8*30}[/tex]
[tex]v=24.25[/tex]
Generally the equation for Kinetic Energy is is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}*100*(24.25)^2\\[/tex]
[tex]K.E=29403.125J[/tex]
[tex]K.E=29.403125J[/tex]
The man will cover the horizontal distance with the velocity of efflux. And the required kinetic energy associated with the motion of man is
29.40 J.
When an object is in motion along a linear or curvilinear path, having some magnitude of velocity, then the energy associated with the object in that state is known as kinetic energy.
Given data -
The mass of the man is, m = 100 kg.
The height of the cliff is, H = 50 m.
The distance from the water surface is, h' = 20 m.
We can use the concept of efflux. When water comes out to some horizontal distance, then the velocity associated with the flow is known as the velocity of efflux.
Then mathematical expression for the velocity of efflux is,
[tex]v=\sqrt{2g(H-h')}[/tex]
Here, g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2 \times 9.8(50-20)}\\\\v = 24.25 \;\rm m/s[/tex]
Now, the expression for the kinetic energy of man is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
Solving as,
[tex]KE = \dfrac{1}{2} \times 100 \times 24.25^{2}\\\\KE = 29.40 \;\rm J[/tex]
Thus, we can conclude that the kinetic energy of man diving out at 20 m from water is 29.40 J.
Learn more about kinetic energy here:
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