1 First, the ball was hit at an initial velocity of 40m/s along the direction where the starting point made an angle of 30 degrees with the horizon, and flew at a constant speed. Therefore, the height of the ball can reach 15m, and the required time: let the height h= 15, substitute it into the relation 20t-5t 2 = 15, and solve the quadratic equation in one variable to get t 1= 1, so the time is 3. Flight altitude15m.

2. The flying height can reach 20 meters, let the height h=20, substitute the relation 20t-5t 2 = 20, and solve the unary quadratic equation to get t 1=t2=2, so the flying time is 2S when the flying height reaches 20 meters.

3. The flying height of the ball cannot reach 20.5 meters.

It takes time from flying out to landing. Let the height h=0 and substitute it into the relation 20t-5t 2 = 0. Solve the quadratic equation of one variable and get t=4, so it takes 4S for the ball to fly out and land.

Hope to adopt.