Answer:
2.83227 m/s
Explanation:
K = Kinetic energy
P = Potential energy
v = Final velocity
u = Initial velocity
g = Acceleration due to gravity = 9.81 m/s²
h = Height of the bar
Here the energies of the system are conserved
[tex]K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv^2+mgh=\frac{1}{2}mu^2+0\\\Rightarrow gh=\frac{v^2-u^2}{2}\\\Rightarrow gh=\frac{8.27^2-0.742^2}{2}\\\Rightarrow gh=33.921168[/tex]
The same can be applied to the second pole vaulter
[tex]gh=\frac{v^2-u^2}{2}\\\Rightarrow u=\sqrt{v^2-2gh}\\\Rightarrow u=\sqrt{8.71^2-2\times 33.921168}\\\Rightarrow u=2.83227\ m/s[/tex]
The speed at which the second vaulter clears the bar is 2.83227 m/s
