Question

A student is bouncing on a trampoline. At her highest point, her feet are 95 cm above the trampoline. When she lands, the trampoline sags 25 cm before propelling her back up.

Answer 1

The time she spent in contact with the trampoline is 0.1619 seconds.

The objective of this question is to determine for how long she is in contact with the trampoline?

The conservational force is exerted on the trampoline is equivalent to the sum of the potential energy at the highest point and the elastic energy at the lowest point.

∴

From the given information;

$\mathbf{mg \dfrac{(95 +25)}{100} = \dfrac{1}{2}k ( \dfrac{25}{100})^2}$

$\mathbf{m\times g\times 1.2 = k(0.03125)}$

$\mathbf{m\times 9.8 \times 1.2 = k(0.03125)}$

$\mathbf{\dfrac{m}{k} = \dfrac{(0.03125)}{9.8 \times 1.2}}$

$\mathbf{\dfrac{m}{k} = 0.002657}$

Using the Simple Harmonic Motion to determine the time at which she is in contact with the trampoline, we have;

$\mathbf{t= \dfrac{T}{2}}$

$\mathbf{t= \dfrac{2 \pi}{2} \sqrt{\dfrac{m}{k}}}$

where;

• $\mathbf{\dfrac{m}{k} = 0.002657}$

$\mathbf{t= \dfrac{2 \pi}{2} \times \sqrt{0.002657}}}$

$\mathbf{t=0.1619\ sec}}$

Learn more about the time period of a Simple Harmonic Motion here:

brainly.com/question/17315536