As shown in the figure, it is a photo taken by stroboscopic photography to study the motion law of objects in flat throwing. Use a piece of paper with small squares on it to record the trajectory.

The time interval between two adjacent points is t, and the vertical direction is 2L-L=gT2, so:

T=△yg=Lg= 1.25× 10？ 29.8 seconds ≈0.036 seconds

Horizontal direction: v0 = 2lt =1.25x10? 2×2m0.036s≈0.7m/s

According to the law of uniform linear motion, the vertical velocity of point B is:

vby = hac2T = 3L2T = 3× 1.25× 10？ 2 m2× 0.036 sec = 0.52 m/sec.

Therefore, the time from the throwing point to the point B is:

t = vbyg = 0.52m/s 10m/S2 = 0.052s

Then the time from the throwing point to the point A is:

△T = T-T = 0.052s-0.036s = 0.0 16s

Therefore, the coordinates of the throwing point are: y =12g (△ t) 2 =12× 9.8× (0.016) 2 =1.25×10? 3m=0. 125cm

x = v0t = 0.7×0.0 16 = 0.0 1 12m = 1. 12cm

So the answer is: 0.036, (1. 12cm, 0. 125cm).