9) A 40 kg gymnast is practicing her routine. On her final jump, she reaches a height of 2.0 meters before landing on the gym mat. The mat stops the gymnast in 0.05 seconds. a. What is the magnitude and direction of the impulse required to stop the gymnast? b. What is the average force the mat exerts on the gymnast? c. Our gymnast decides to practice her routine at home on the driveway and sprains her ankle. The driveway stops the gymnast in 0.01 seconds. Explain why the gymnast injured herself at home (use calculations to justify your answer):

The speed of a falling object can be found using constant acceleration kinematics or conservation of energy. When the gymnast lands on the gym mat, the mat exerts a force on the gymnast that stops their fall. The impulse of this force is equal to the change in momentum, or the force times the amount of time.

a. First, we need to find the gymnast's speed when she lands on the mat. We can do this using kinematics or energy. Taking down to be positive, if we use kinematics:

v² = u² + 2as

v² = (0 m/s)² + 2 (9.8 m/s²) (2.0 m)

v = 6.26 m/s

Alternatively, if we use energy:

PE = KE

mgh = ½ mv²

v² = 2gh

v² = 2 (9.8 m/s²) (2.0 m)

v = 6.26 m/s

Next, the impulse is equal to the change in momentum.

J = Δp

J = mΔv

J = (40 kg) (0 m/s − 6.26 m/s)

J = -250 kg m/s

The impulse is 250 kg m/s upwards.

b. The impulse is equal to the force times the time.

J = FΔt

250 kg m/s = F (0.05 s)

F = 5000 N

c. This time, the gymnast comes to a stop in 0.01 seconds.

J = FΔt

250 kg m/s = F (0.01 s)

F = 25,000 N

The force is five times higher when the gymnast lands on the driveway. This higher force causes the gymnast to sprain her ankle.