Flight time of the first bomb t1= 2hg = 2× 200010s = 20s;

The vertical speed when landing is v1= gt1=10× 20m/s = 200m/s.

It is easy to know from mathematical knowledge that the included angle θ is 45.

(2) The horizontal distance from the foot of the mountain when the plane dropped the first bomb is: x1= v0t1= 200× 20m = 4000m.

The flight time of the second bomb t2=2(H? h)g=2×(2000？ 720) 10s= 16s

The horizontal flight distance is: x2=v0t2=200× 16m=3200m.

(3) The dropping time interval is △t=x 1+ 1200? x2v=4000+ 1200？ 3200200s= 10s

A:

(1) When the first bomb reaches the target A, the included angle θ between its speed direction and the horizontal ground is 45 degrees.

(2) The horizontal distances of the two dropped bombs flying in the air are 4000m and 3200m respectively. They are.

(3) The time interval △t between two bombs is 10s. ..