A 51.5-kg athlete leaps straight up into the air from a trampoline with an initial speed of 7.2 m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? This answer has not been graded yet. (b) Select the height at which the athlete's speed is 7.2 m/s as y = 0. What is her kinetic energy at this point? J What is the gravitational potential energy associated with the athlete? J (c) What is her kinetic energy at maximum height? J What is the gravitational potential energy associated with the athlete? J (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. m (e) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

(a) The interacting objects are the athlete and the trampoline. The trampoline would store potential energy and transfer to the athlete's kinetic energy.

(b) Her kinetic energy at y = 0

[tex]E_{k0} = \frac{mv^2}{2} = \frac{51.5*7.2^2}{2} = 1334.88J[/tex]

Her gravitational energy at y = 0

[tex]E_{g0} = mgy = 51.5*9.81*0 = 0 J[/tex]

(c)At maximum height, her speed would be 0, and so is her kinetic energy

[tex]E_{kmax} = \frac{51.5*0^2}{2} = 0 J[/tex]

let [tex]y_{max}[/tex] be the maximum height that she reaches, then her potential energy at this point is

[tex]E_{gmax} = mgy_{max} = 51.5*9.81*y_{max} = 505.215y_{max}J[/tex]

(d) By the law of energy conservation at y = 0 and y max:

[tex]E_{k0} + E_{g0} = E_{kmax} + E_{gmax}[/tex]

[tex]1334.88 + 0 = 0 + 505.215y_{max}[/tex]

[tex]y_{max} = 1334.88/505.215 = 2.64 m[/tex]

(e) At half maximum way y2 = 2.64/2 = 1.32 m

[tex]E_{k0} + E_{g0} = E_{k2} + E_{g2}[/tex]

[tex]1334.88 + 0 = \frac{51.5*v_2^2}{2} + 51.5*9.81*1.32[/tex]

[tex]\frac{51.5*v_2^2}{2} = 1334.88 - 667.44 = 667.44[/tex]

[tex]v_2^2 = 25.92[/tex]

[tex]v = 5.1 m/s[/tex]