Question

A 53.0-kg athlete leaps straight up into the air from a
trampoline with an initial speed of 7.5 m/s. The goal of this
problem is to find the maximum height she attains and her speed at
half maximum height.

a. Select the height at which the athlete's speed is 7.5 m/s
as

y = 0. What is her kinetic energy at this point?

What is the gravitational potential energy associated with the
athlete?

b. What is her kinetic energy at maximum height?

What is the gravitational potential energy associated with the
athlete?

c. Write a general equation for energy conservation in this
case and solve for the maximum height. Substitute and obtain a
numerical answer.

d. Write a general equation for energy conservation in this
case and solve for half the maximum height. Substitute and obtain a
numerical answer.

Answer #1

a)

K.E = 1/2mv^{2}

K.E = 1/2 * 53 * 7.5^{2}

K.E = 1490.625 J

and

Gravitational potential energy = 0 J

-----------------

b)

K.E ( max height) = 0 J

Gravitational potential energy = 1490.625 J

--------------------------

c)

1/2mv1^{2} + mgy1 = 1/2mv2^{2} + mgy2

y2 = y1 + (( v1^{2} - v2^{2} ) / 2g )

y2 = 0 + ((7.5^{2} - 0) / 2 * 9.8)

y2 = 2.87 m

----------------------------------------------------

(d)

just use kinematics

v2 = sqrt ( v1^{2} + 2g ( y1 - y2))

v2 = sqrt ( 7.5^{2} - 2 * 9.8 * 0.5
*2.87)

v2 = 5.3 m/s

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