In military training, soldiers jump off the wall so at a speed vo, as shown in the figure, and then use their pedals to the wall again, so that their bodies become vertical and upward and continue to climb the wall. Soldiers do oblique upward throwing after take-off, and the horizontal speed is VX = v0cos θ ... ①.

The vertical velocity is vy=v0sinθ…②...② ② ... ②.

The time from jumping to hitting the wall is:

t=s0vx…③

The vertical height in this time is: h=vyt? 12gt2…④

After the soldier touches the wall, let the horizontal force of kicking the wall with the action time of t' be f, which is obtained from the momentum theorem in the horizontal direction:

-Ft′=-mvx⑤

The friction in the vertical direction is f=μF⑥.

During the interaction time with the wall, the soldier makes a uniform acceleration linear motion with an acceleration of a, which is obtained by Newton's second law:

f-mg=ma⑦

The vertical velocity of interaction between soldiers and walls is: v ′ y = vy+at 8.

After that, the soldiers do uniform deceleration linear motion, and the rising height is:

h′= v′2y2g⑨

The total height of soldiers rising is: H=h+h'( 10).

Use the above formula 10 to finish at the same time, and the angle is θ = Arctanv 20gs0.

Answer: The take-off angle that can make the human center of gravity have the largest total increment is arctanv20gs0. ..