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podryga [215]
2 years ago
12

A father places his daughter in a swing that is 0.60\,\text{m}0.60m0, point, 60, start text, m, end text above ground. Then he r

aises the swing to a height of 1.3\,\text{m}1.3m1, point, 3, start text, m, end text and lets go. The girl and the swing have a combined mass of 14\,\text{kg}14kg14, start text, k, g, end text. Assume friction is negligible and use g = 9.8\,\text{m/s}^2g=9.8m/s
2
g, equals, 9, point, 8, start text, m, slash, s, end text, squared.
Calculate the girl’s fastest speed.
Physics
1 answer:
schepotkina [342]2 years ago
5 0

This question involves the concepts of the law of conservation of energy and kinetic energy.

The girl's fastest speed is "3.7 m/s".

According to the law of conservation of energy, the girl will have the fastest speed at mean position, which will be calculated as follows:

Loss in Potential Energy = Gain in Kinetic Energy

mg\Delta h=\frac{1}{2}mv^2\\\\v=\sqrt{2g\Delta h}

where,

v = maximum speed = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = change in height = 1.3 m - 0.6 m = 0.7 m

Therefore,

v=\sqrt{2(9.81\ m/s^2)(0.7\ m)}

<u>v = 3.7 m/s</u>

<u></u>

Learn more about the Law of Conservation of Energy here:

brainly.com/question/381281?referrer=searchResults

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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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