A student is bouncing on a trampoline. At her highest point, her feet are 45 cm

Question

A student is bouncing on a trampoline. At her highest point, her feet are 45 cm above the trampoline. When she lands, the trampoline sags 25 cm before propelling her back up.

A) For how long is she in contact with the trampoline? The answers are NOT 0.53 s, 0.17 s, or 0.26 s

We must use Conservation of Energy:

given at the highest point, ho = 0.45 m

when she lands, the trampoline sags by xo = 25 cm

Then,

E = m*g*ho = 0.5*k*xo2-m*g*xo where k is spring constant of the trampoline

0.5k*xo2-mg*xo = mg*ho(a)

So, period of oscilation of the trampoline: T = 2*pi*sqrt(m/k) (b)

Solve for T, period from the above equations (a) & (b)

Explanation / Answer

Given that,

h = 45 cm = 0.45 m

trampoline sags by, x = 25 cm = 0.25 m

From law of energy conservation,

mgh = (1/2)kx^2 - mgx

m*9.8*0.45 + m*9.8*0.25 = (1/2)k*(0.25)^2

m / k = 0.0045553

period of oscilation of the trampoline,

T = 2*pi*sqrt (m / k)

Put the value of m/k,

T = 2*3.14*sqrt (0.0045553)

T = 0.424 s

She is in contact with trampoline for 0.424 s

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