[tex]U=T_{2} -T_{1} \\-m*g*h_{max}+\frac{1}{2}*k*(x_{1}^{2} -x_{2}^{2})=0-0\\ -0.19*9.81*h_{max}+\frac{1}{2}*650*(0.16^{2} -0^{2})=0\\h_{max}=4.464 m[/tex]
The maximum height reached by the baseball is 4.31 m.
Given that the compression/ displacement of the spring is d = 0.16 m.
The mass of the baseball is m = 0.19 kg
and the spring constant is k = 650 N/m
The gravitational potential energy at the compressed position is zero.
According to the law of conservation of energy, the total energy of the system remains conserved under the influence of conservative forces like spring force.
Let the final height from the bottom be h.
Then,
[tex]KE_{final}+PE{final}=KE_{initial}+PE_{initial}+PE_{spring}[/tex]
the initial kinetic energy is zero, also at the highest point, the final kinetic energy is also zero, therefore:
[tex]PE_{final}=PE_{spring}\\\\mgh=\frac{1}{2}kd^2\\\\h=\frac{kd^2}{2mg}\\\\h=\frac{650\times(0.16)^2}{2\times0.19\times9.8}\\\\h=4.47\;m[/tex]
Now the maximum height from the mean position of the spring will be:
H = h-d = (4.47 - 0.16) m
H = 4.31 m
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