Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
[tex]KE_i+PE_i =KE_f+PE_f[/tex]
or
[tex]\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh[/tex]
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump:
[tex]u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s[/tex]
