Answer:
2856.96 J
0
0
[tex]\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f[/tex]
6.78822 m/s
Explanation:
[tex]v_i[/tex] = Initial velocity = 9.6 m/s
g = Acceleration due to gravity = 9.81 m/s²
h = Height
The athlete only interacts with the gravitational potential energy. Air resistance is neglected.
At height y = 0
Kinetic energy
[tex]K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J[/tex]
At height y = 0 the potential energy is 0 as
[tex]P=mgy\\\Rightarrow P=mg0=0[/tex]
At maximum height her velocity becomes 0 so the kinetic energy becomes zero.
As the the potential and kinetic energy are conserved
The general equation
[tex]K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f[/tex]
Half of maximum height
[tex]\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}[/tex]
[tex]h_i=\frac{v_i^2}{2g}[/tex]
[tex]v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s[/tex]
The velocity of the athlete at half the maximum height is 6.78822 m/s
