Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$

K = (1/2)mv^{2} and U = mgh

Law of conservation of momentum:

$\overline{)\begin{array}{rcl}{\mathbf{p}}_{\mathbf{i}}& {\mathbf{=}}& {\mathbf{p}}_{\mathbf{f}}\\ {\mathbf{p}}& {\mathbf{=}}& \mathbf{m}\mathbf{v}\end{array}}$

**a)**

Let v_{2} be the speed of the girl before immediately before she grabs the box.

From the law of conservation of energy:

$\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{2}}\overline{){\mathbf{m}}_{\mathbf{1}}}{{\mathbf{v}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}\mathbf{0}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\overline{){\mathbf{m}}_{\mathbf{1}}}{{\mathbf{v}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}\overline{){\mathbf{m}}_{\mathbf{1}}}\mathbf{g}\mathbf{h}\\ {{\mathbf{v}}_{\mathbf{1}}}^{\mathbf{2}}& \mathbf{=}& {{\mathbf{v}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{g}\mathbf{h}\\ {\mathbf{v}}_{\mathbf{2}}& \mathbf{=}& \sqrt{{{\mathbf{v}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{g}\mathbf{h}}\\ & \mathbf{=}& \sqrt{\mathbf{8}\mathbf{.}{\mathbf{0}}^{\mathbf{2}}\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{81}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{)}}\end{array}$

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v_{1 }= 8.0 meters per second. At height h =2.0 meters above the trampoline, the girl grabs a box of mass m_{2 }= 15 kilograms.

For this problem, use g = 9.8 meters per second for the magnitude of the acceleration due to gravity.

a)What is the speed of the girl immediately before she grabs the box?

b)What is the speed of the girl immediately after she grabs the box?

c)What is the maximum height that the girl (with box) reaches? Measure with respect to the top of the trampoline

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What scientific concept do you need to know in order to solve this problem?

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