Asked Oct 20, 2019

An olympic gymnast jumps up and down on a trampoline. If the gymnast's feet leave the trampoline at a height of 1.0 m above the ground and reach a maximum height of 6.0 m above the ground, at what speed was the gymnast launched from the trampoline?


Expert Answer

Step 1

According to the principle of conservation of energy, the sum of the change in potential energy of the gymnast and the change in his kinetic energy will be equal to zero.


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AUAK 0 Here, AU is the change in potential energy and AK is the change in kinetic energy The equation for the change in potential energy of the gymnast is given by (1) AU mg (h,-h) (2) Here, is the mass of the gymnast, g is the acceleration due to gravity, h, is the final position of gymnast and h is the initial position The equation for the change in kinetic energy of the gymnast is given by 1 AK m-) is the speed of gymnast at h, and v, is the speed of gymnast at h Here, Take the final position to be the maximum height reached by the gymnast. At maximum height, his velocity will be zero. Substitute 0 for v, in the above equation xm(0-) AK mv (3)

Step 2

Put equations (2) and (3)...


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mg (h,-h) =0 s(,-) 2 - 2g (h -) ) ,2g(h,-h Substitute the numerical values in the above equation. 2(9.80 m/s2)(6.0 m -1.0 m) 9.9 m/s


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