Question

A student is bouncing on a trampoline. At her highest point, her feet are $55cm$above the trampoline. When she lands, the trampoline sags$15cm$before propelling her back up. For how long is she in contact with the trampoline?

Short answer

She is in contact with the trampoline for$t_c=T=0.134s$

Step-by-step solution

Given data.

Given

• $d=55.0cm$
  
• $2A=15cm$
  

Required:$T$

Solution

$x=0$The 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

$v_{max}=A\omega =\frac{2\mathrm{\pi A}}{T}\Rightarrow \left(1\right)$

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

$\frac{1}{2}m{v^2}_{max}=mgh\Rightarrow v_{max}=\sqrt{2gh}\Rightarrow \left(3\right)$

We assume that student go down $7.5cm$while she is in contact with trampoline so $x=0$after going down $7.5cm$so $h$is given by

$h=55+7.5=62.5cm=0.625m$

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

$v_{max}=\sqrt{2\times 0.625\times 9.8}=3.5m/s$

substitution in $\left(1\right)$yields

$T=\frac{2\mathrm{\pi }\times 0.075}{3.5}=0.134S$

$v_{max}=\sqrt{2\times 0.625\times 9.8}=3.5m/s$