In order to solve this question we must first solve for how long the ball is in the air:

`v_2-v_1 = at -gt t=(v_2-v_1)/a`

`t=(0-20)/-9.81=2.04s`

At 2.04s the ball has reached its maximum height

Next we must calculate its maximum height:

`d=((v_2+v_1)/2)t=(10)(2.04)=20.4m`

After 2.04s the ball has travelled 20.4m vertically.

Now we need to know how long it takes the ball to fall back to the throwers hand. Remember, that `v_1` is now 0 (starting from rest at max height):

`d=v_1t+1/2at^2-gtt=sqrt(2d/a)=sqrt(2(20.4)/9.81)=2.04s`

As expected, the ball takes the same amount of time to fall as it does to raise. Therefore, the total time spent in the air is 2.04+2.04=4.08s.

The last step is to determine how far the bus has travelled in 4.08s:

`d=vt=(30)(4.08)=122.4m`

Although, it is important to note that the ball would not return to the thrower's hand, but would fall to the ground 122.4m behind the thrower at the bus' position when the ball was thrown.

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