A few days ago
James

Intriguing Physics Problem. Can you help!?

Please explain if you can. Thanks!!

A gymnast practices two dismounts form 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 form the same height but with and initial downward velocity of -3.0m/s. How do the final velocities of the gymnast as she reaches the ground differ? What is her acceleration in each case?

Top 1 Answers
A few days ago
Anonymous

Favorite Answer

Her acceleration in all cases is ‘-g’.

In the first case, she is initially going upwards at 4m/s, but she will decelarate to 0 and then accelerate to -4m/s as she passes the bar. This is because the ‘g’ is acting over the same distance as she goes up and then as she goes down.

You know that v^2 = u^2 + 2as

‘u’ in the first case is -4, in the second case is -3

‘a’ and ‘s’ are the same in both cases (-g and -h the height of the bar)

v = root( (-3)^2 + 2gh )

w = root( (-4)^2 + 2gh )

w – v = (the difference of the two expressions)

You can see that the final velocity in the first case will have a larger magnitude, but the exact difference depends on the height.

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