Questions: The center of mass moves with constant acceleration when constant forces act on the object. A gymnast jumps straight up, with her center of mass rising at 1.80 m/s at the top of her jump. Ignore the effects of air resistance, and assume the initial height of her center of mass is zero. (Ignore the effects of air resistance, and assume the initial height of her center of mass is zero.) (a) 0.100 s (b) 0.200 s (c) 0.300 s (d) 0.500 s

Solution Steps

We are given:

• The initial velocity at the top of the jump, \( v = 1.80 \, \text{m/s} \)
• The acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
• The initial height of the center of mass is zero.

At the top of the jump, the gymnast's velocity will be zero. We can use the kinematic equation: \[ v_f = v_i + at \] where:

• \( v_f = 0 \, \text{m/s} \) (final velocity at the top)
• \( v_i = 1.80 \, \text{m/s} \) (initial velocity)
• \( a = -9.81 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it is downward)
• \( t \) is the time to reach the top.

Rearranging the equation to solve for \( t \): \[ 0 = 1.80 - 9.81t \] \[ 9.81t = 1.80 \] \[ t = \frac{1.80}{9.81} \] \[ t \approx 0.1834 \, \text{s} \]

The calculated time \( t \approx 0.1834 \, \text{s} \) is closest to option (b) 0.200 s.

Final Answer