Intriguing Physics Problem. Can you help!?
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?
Favorite Answer
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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