Question

A gymnast jumps straight up, with her center of mass moving at 4.73 m/s as she leaves the ground. How high above this point is her center of mass at the following times? (Ignore the effects of air resistance, and assume the initial height of her center of mass is at y = 0.)

Answer 1

Answer:

At any time t, her altitud will be given by the following law:

$y=0 + 4.73\frac{m}{s}\cdot t - \frac{1}{2} g\cdot t^{2}$

Explanation:

We can consider the gymnast as a point particle in the position of her center of mass. The law of movement will be:

$y=0 + 4.73\frac{m}{s}\cdot t - \frac{1}{2} g\cdot t^{2}$

where g is gravity's acceleration which we can take as $g=9.8\frac{m}{s^{2} }$. We add 0 in the last member just to note that at $t=0$ the vertical position is 0. If we knew that she was moving at $4.73\frac{m}{s}$ at $1m$, we would have needed to put 1m in the place of 0.