Question

4. A 60 kg gymnast falls from the high bar to a mat. The gymnast's vertical velocity at the instant of bar release is 0.0 m/s and his center of gravity is 2.5 m above the mat. When the gymnast first contacts the mat, his center of gravity is 1.0 m above the mat. At this instant of initial contact, what is the gymnast's vertical velocity?

5. When the gymnast finally comes to a stop, his center of gravity is only 25 cm above the mat. What was the average vertical reaction force exerted on the gymnast by the mat during his landing?

Answer 1

4) use conservation of energy

gain in kinetic energy = loss of potentail energy

(1/2)*m*v^2 = m*g*(h1-h2)

v = sqrt(2*g*(h1-h2))

= sqrt(2*9.8*(2.5-1))

= 5.42 m/s <<<<<<<<<<<<<<-------------------Answer

5) let a is the constant acceleration of the gymnast.

a = (vf^2 - vi^2)/(2*a*d)

= (0^2 - 5.42^2)/(2*0.75)

= -19.6 m/s^2

|a| = 19.6 m/s^2

Fnet = N - m*g

m*a = N - m*g

==> N = m*(g+a)

= 60*(9.8 + 19.6)

= 1764 N <<<<<<<<<<<<<<-------------------Answer