Question

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?

Answer 1

The working formula to analyze this problem is Vf^2 - Vo^2 = 2gs where Vf = velocity at which the gymnast hits the ground Vo = initial velocity of the gymnast (given as 4 m/s and -3 m/s) g = acceleration due to gravity = 9.8 m/sec^2 (constant) s = height at which gymnast starts her dismount For the first dismount, Vf^2 - (4)^2 = 2(9.8)(s) Vf^2 = 16 + 19.6s and for the second dismount, Vf^2 - (-3)^2 = 2(9.8)(s) Vf^2 = 9 + 1.6s Since "s" is the same for both dismounts, then her final velocity in the first dismount is higher than that of her second dismount.