Please help me answer this hard physic problem and its components A 48.5-kg athl

Question

Please help me answer this hard physic problem and its components

A 48.5-kg athlete leaps straight up into the air from a trampoline with an initial speed of 7.1 m/s. The goal of this problem is to find the maximum height she attains an her speed at half maximum height. What are the interacting objects and how do they interact? This answer has not been graded yet. Select the height at which the athlete's speed is 7.1 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 What is her kinetic energy at maximum height? J What is the gravitational potential energy associated with the athlete? J Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. m Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer. m/s

Explanation / Answer

(b)

at v = 7.1 m/s

h = 0

KE = (1/2)*m*v^2

KE = (1/2)*48.5*(7.1)^2 = 1222.44 J

gravitational PE = m*g*h

(h = 0)

so, PE = 0

(c)

at maximum height , v = 0

so, KE = 0

gravitaional PE = m*g*hmax

so first we need to calculate hmax.

apply third equation of motion,

v^2 = u^2 + 2*a*hmax

0 = u^2 - 2*g*hmax

hmax = u^2 / 2a

hmax = (7.1)^2 / 2*9.8

hmax = 2.57 m

PEg = 48.5*9.8*2.57

PEg = 1222.44 J

(d)

general equation for energy conservation,

KEi + PEi = KEf + PEf

(1/2)*m*v^2 + 0 = 0 + m*g*hmax

hmax = v^2 / 2*g = 2.57 m

(e)

let v = velocity at half the maximum height.

h = 2.57 / 2 = 1.285 m

general equation for energy conservation,

KEi + PEi = KEf + PEf

(1/2)*m*u^2 + 0 = (1/2)*m*v^2 + m*g*h

(1/2)*(7.1)^2 = (1/2)*v^2 + 9.8*(1.285)

v = 5.02 m/s

answer

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