(1) In the process of putting the player at the lowest point, mechanical energy is conserved, which is obtained by the law of conservation of mechanical energy:

mgl( 1-cosα)= 12mv2，

Solution: v=2gl( 1? cosα)=2× 10×2×( 1？ cos 53)= 4m/s；

At the lowest point, it is obtained from Newton's second law: F-mg=mv2l,

Substituting the data, the result is: F= 1080N.

According to Newton's third law, people's pulling force on the rope is f' = f =1080 n;

(2) When the player swings to the highest point on the right, let go and the player will fall freely. For the whole process, we can get from the kinetic energy theorem:

mg(H-lcosα)-Fd=0

Solution: d=mg(H? cosα)F=600×(3？ 2×cos53 )700m= 1.02m

(3) The player starts to do flat throwing after letting go from the lowest point.

In the horizontal direction: x=vt,

Vertical direction: H-l= 12gt2,

Question 1 Yes: v=2gl( 1? cosα)

Simultaneous: x=2l(H? l)( 1？ cos53)

According to mathematical knowledge, when l=H-l, x has the maximum value, then l =12h =1.5m.

Answer: (1) When the player swings to the lowest point, the tension on the rope is 1080N. ..

(2) When the player reaches the highest point on the right, let go and the player will fall freely. The player's falling depth d is1.02m; ;

(3) If the player is required to let go when the swing reaches the lowest point, and the horizontal distance from the floating platform is the largest, the player should hold the rope with his hand at the distance o1.5m..