Question

With what minimum speed must athlete leave the ground in order to lift his center of mass 1.80 m and cross the bar with a speed of 0.65 m/s ?

Answer 1

Answer:

The minimum velocity is V=5.98[m/s]

Explanation:

In this problem the athlete jumps by printing a force to raise his mass at an initial speed, which we must determine. These two initial amounts of energy that we have, the kinetic energy (velocity with which it jumps) and the work done (mass of the athlete by the jumped distance) must be equal to the kinetic energy when its velocity is equal to 0.65[m/s]. By raising this equation, the initial speed can be cleared.

We have to keep in mind that the weight is acting down and the movement is one up so the work done by the force is negative

Therefore we have:

$Ek1+W=Ek2\\where\\Ek1= initial kinetic energy\\W= work done\\Ek2=final kinetic energy$

$\frac{1}{2} *m*v_{0} ^{2} - (m*g*h)=\frac{1}{2} *m*v_{f} ^{2}$

Reorganizing each of the members of the equation and clearing the final velocity.

$v_{0} =\sqrt{(v_{f})^{2} +2*g*h} \\v_{0} =\sqrt{(0.65)^{2} +2*9.81*1.8} \\v_{0} =5.98[m/s)$