An athlete jumping vertically on a trampoline leaveles the surface with a velocity of `8.5 m//s` upwards. What maximum height does the reach?

Options

A. `10 m`
B. `2.5 m`
C. `5.0 m`
D. `0.50 m`

Answer

C

Explanation

Once the athlete leves the surface the tranpoline , only a conservative force (het weight) acts on her .Therefore the , mmechanical energy of the athlete - earth system is constent during her flight `k_(f) +U_(f) = k_(i) +U_(i))` .Taking the `y = 0` at the surface of the tranpoline.`U_(i) = mgv_(i)= 0` .Also , her speed when she rwaches maximum height ,zero , or `k_(f) = 0` This levels as with `U_(f) = k_(i)`, or` mgv_(max) = (1)/(2) mv_(i)^(2)` , which gives the maximum height as
`y_(max) = (v_(i)^(2))/(2g)= ((10.0m//s)^(2))/(2(10.0m//s^(2))) = 5.0m`