Question

Zoe, a 55 kg gymnast is performing a vault. Her center of mass (COM) is 1.0 m above the ground as she runs toward the vault apparatus. The instant she before she jumps into the air, her running velocity is 5 m/s. Her mechanical energies due to her approach velocity and due to her COM position during the approach are completely converted to her total mechanical energy while she is in the air. (a) What is the peak height of Zoe’s COM while she is in the air? (b) What is Zoe’s vertical velocity the instant she lands upright on a mat 0.6 m thick (COM height = 1.6 m from the ground)? (c) The mat compresses to a maximum depth of 0.2 m during Zoe’s landing. At the point of maximum compression of the mat, her vertical velocity reaches zero and her COM is 1.2 m above the ground. How much work was done by the mat? (d) What average force did the mat exert on Zoe during the impact? (e) What was Zoe’s change in momentum while she was in contact with the mat? (f) How long did it take Zoe to come to rest while she was in contact with the mat?

Answer 1

Givenv that

center of mass (COM) h1 =  1.0 m

gymnast mass m = 55 kg

intial velocity is v= 5 m/s

Part(a)

Using conservation of energy

mgh = 1/2mv^2 + mgh1

55*9.8*h = 0.5*55*5^2 + 55*9.8*1

h = 2.28 m

Part(b)

now after reaching at peak height

mgh = 1/2mv1^2 + mgh1

here h1 = 1.6 m

h = 2.28 m

55*9.8*2.28 = 0.5*v1^2 +55*9.8*1.6

v1 = 3.65 m/s

Part(c)

We know that

work energy theorem

mgh2 -w = -1/2 mv1^2

here h2 = 0.2 m

55*9.8*0.2 - w = - 0.5*55*3.65^2

w = 474.17 J

Part(d)

mgh2 -w = -1/2 mv1^2

mgh2 -F.h2 = -1/2 mv1^2

55*9.8*0.2 - F(0.2) = - 0.5*55*3.65^2

F = 2370.84 N

Good Luck !!!