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IgorLugansk [536]

3 years ago

13

You have a friend who is entering a skateboarding competition in Los Angeles. She wants to take a running start and then jump on

to her 5 kg stationary skateboard. She will then roll along a 2 meter section of level track and go up a curved concrete ramp. She wants to reach a maximum height of at least 3 meters above where she started. She can run with a maximum speed of 8 m/s and has a mass of 50 kg. Will this trick work?

1 answer:

ahrayia [7]3 years ago

5 0

Answer

given,

mass skateboard = 5 Kg

length on the horizontal road = 2 m

maximum speed she can run= 8 m/s

mass of girl = 50 Kg

conservation of momentum

M V = (M+m)v

50 x 8 = (50 + 5)v

400 = (55)v

v = 7.27 m/s

using conservation of momentum

$m g h = \dfrac{1}{2} m v^2$

$h = \dfrac{v^2}{2g}$

$h = \dfrac{7.27^2}{2\times 9.8}$

h = 2.69 m

no, she will not be able to reach the height of 3 m

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The given question is incomplete. The complete question is as follows.

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U = $(\frac{1}{2}m_{p}v^{2}_{p})$

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