The maximum height of the child above the ground is 1.3 m
Explanation:
We can solve this problem by using the law of conservation of energy. In fact, the mechanical energy of the child (the sum of his gravitational potential energy + his kinetic energy) must be conserved during the motion on the swing. Therefore we can write:
[tex]U_i +K_i = U_f + K_f[/tex]
where
[tex]U_i[/tex] is the potential energy at the bottom
[tex]K_i[/tex] is the kinetic energy at the bottom
[tex]U_f[/tex] is the potential energy at the top
[tex]K_f[/tex] is the kinetic energy at the top
The equation can be also rewritten as
[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]
where:
m is the mass of the child
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]h_i = 0.40[/tex] is the height of the child at the lowest point
u = 4.2 m/s is speed of the child at the lowest point (the maximum speed occurs at the lowest point)
[tex]h_f[/tex] is the maximum height reached by the skateboarder
v = 0 is the speed at the maximum height
And solving for [tex]h_f[/tex], we find:
[tex]\frac{1}{2}mu^2 + mgh_i = mgh_f\\h_f = h_i + \frac{u^2}{2g}=0.40+\frac{(4.2)^2}{2(9.8)}=1.3 m[/tex]
Learn more about kinetic energy and potential energy here:
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