Agymnast 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?

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.

The answer to this is decibel

No because scientist are still changing there mines today so they are not correct to this day