From the balcony of a 20 meter high building, a small one is thrown vertically at an initial velocit

I choose ACD

There are 3 situations in this problem, one is to move upwards 15m above the throwing point, and the other is to move downward 15m above the throwing point. The third is to fall 15m below the throwing point.

1, by the formula: v0 2-v 2 = 2gh

So: v=10m s

By velocity formula: v=v0-gt

t=(20-10)/10=1s

2. After 1s upward, after 15 displacements, there is still 5m, and the time t2 needs to reach the highest: h=gt2 2 2=5, t2=1s

Then it takes 1s to fall back to 15m, so the time t=1+1+1=3s3 to reach the upper 15m for the second time, and the time to reach the highest point t1=v0 g=20 10=2s If it is 15m below the throwing point, it also needs to do free fall H=35m, and the total time required to fall from H=gt2 2 2 t2= 2H g= 7s t=t1+t2=(2+7)s

The throw point is 20m high and 15m away from the throw point, which can be above or below the throw point.

Stipulate that the upward direction is positive.

Above the throw point: then the displacement x = 15 m, the initial velocity vo = 20 m s, and the acceleration a = -g = -10 m s

x=vot+½at²

15=20t+½×10)t²

Solve t1=3s or t2=1s

Below the throw point: then the displacement x=-15m, the initial velocity vo=20ms, and the acceleration a=-g=-10ms

x=vot+½at²

15=20t+½×10)t²

Solve t1=2+ 7 or t2=2- 7 (negative values are rounded), so ac d is correct.

The square of v0 = 2gh has to rise at a height of h0 = 20m, so the height of the ball is greater than 15 meters.

The first point of the rise is the square of v1 - the square of vo = 2gh and v1 = 10m s t19 = (v0-vi) g = 1s

The second point ball drops from the highest point to the inside throw point 15 meters 1 2GT2"square = h0-h to get t2"=1s gt2'=v0 to get t2'=2s so t2=1+2=3s

The third point is fifteen meters below the throwing point 1 2GT3'square = h0 + 15 to get t3'=root7 so t3=t3'+t2'=2 + followed by 7

Choose a c d typing is so tangled!

Solution: The displacement of the ball from the balcony to the highest point is S=V0 2 2g=20 2 20=20m>15m, so there are three cases when the ball moves to 15 meters from the throwing point. Let the direction up be positive and down negative.

1）.When the displacement is 15m, from x=V0t-gt2 2, 15=20t-10t2 2 is obtained, and t=1 or 3 is solved. （2）.

When the displacement is -15m, in the same way, -15=20t-10t2 2, the solution is t=2 + root number 7 or 2-root number 7 (rounded), so choose a, c, d