The speed of the rock at 20 m is 34.3 m/s
Explanation:
We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:
[tex]U_i +K_i = U_f + K_f[/tex]
where :
[tex]U_i[/tex] is the initial potential energy
[tex]K_i[/tex] is the initial kinetic energy
[tex]U_f[/tex] is the final potential energy
[tex]K_f[/tex] is the final kinetic energy
The equation can also be rewritten as follows:
[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]
where:
m = 100 kg is the mass of the rock
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]h_i = 80[/tex] is the initial height
u = 0 is the initial speed (the rock starts at rest)
[tex]h_f = 20 m[/tex] is the final height of the rock
v is the final speed when h = 20 m
And solving for v, we find:
[tex]v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s[/tex]
Learn more about kinetic energy and potential energy here:
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