The athlete must leave the ground with a speed of 6.33m/s.
To calculate the initial speed, with the information of height and final speed, we can use the following expression that uses the kinetic energy and potential energy:
[tex]mgh_1 + \frac{1}{2}mv_1^{2}= mgh_2 + \frac{1}{2}mv_2^{2}[/tex]
Applying the values given by the statement we have:
[tex]9.81\times 0 + \frac{1}{2}v_1^{2} = 9.81 \times 2 + \frac{1}{2}(0.95)^{2}[/tex]
[tex]v_1 = 6.33m/s[/tex]
So, the athlete must leave the ground with a speed of 6.33m/s.
Learn more about kinect and potential energy in: brainly.com/question/14509624
