A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar with an initial upward velocity of + 4.0 m/s. in the second, she releases from the same height but with an initial downward velocity of - 3.0 m/s. what is her acceleration in each case? how do the final velocities of the gymnast as she reaches the ground differ?

Question

Rene Booth

Physics

17 December, 23:49

A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar with an initial upward velocity of + 4.0 m/s. in the second, she releases from the same height but with an initial downward velocity of - 3.0 m/s. what is her acceleration in each case? how do the final velocities of the gymnast as she reaches the ground differ?

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Answers (1)

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Moriah Whitney · Answer 1

Note:

The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).

It is not provided in the question, so the standard height is assumed.

g = 9.8 m/s², acceleration due to gravity.

Note that the velocity and distance are measured as positive upward.

Therefore the floor is at a height of h = - 2.8 m.

First dismount:

u = 4.0 m/s, initial upward velocity.

Let v = the velocity when the gymnast hits the floor.

Then

v² = u² - 2gh

v² = 16 - 2*9.8 * (-2.8) = 70.88

v = 8.42 m/s

Second dismount:

u = - 3.0 m/s

v² = (-3.0) ² - 2*9.8 * (-2.8) = 63.88 m/s

v = 7.99 m/s

The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.

Answer:

First dismount:

Acceleration = 9.8 m/s² downward

Landing velocity = 8.42 m/s downward

Second dismount:

Acceleration = 9.8 m/s² downward

Landing velocity = 7.99 m/s downward

The landing velocities differ by 0.43 m/s.