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Chapter 15: Q.19 (page 415)

A student is bouncing on a trampoline. At her highest point, her feet are 55cmabove the trampoline. When she lands, the trampoline sags15cmbefore propelling her back up. For how long is she in contact with the trampoline?

Short Answer

Expert verified

She is in contact with the trampoline fortc=T=0.134s

Step by step solution

01

Given data.

Given

  • d=55.0cm
  • 2A=15cm

Required:T

02

Solution

x=0The simple harmonic motion begins when student's feet touch trampoline and the time she will be in contact with trampoline is the periodic time which is given by

vmax=A蝇=2蟺础T鈬�(1)

We need to determine vmaxso we sill use conservation's law, At max point potential is max and at x=0kinetic energy is max so

12mv2max=mgh鈬�vmax=2gh鈬�(3)

We assume that student go down 7.5cmwhile she is in contact with trampoline so x=0after going down 7.5cmso his given by

h=55+7.5=62.5cm=0.625m

The max velocity is achieved when x=0so it is given by

vmax=2脳0.625脳9.8=3.5m/s

substitution in (1)yields

T=2蟺脳0.0753.5=0.134S

vmax=2脳0.625脳9.8=3.5m/s

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