SOLUTION: A twirler throws a baton with an initial upward velocity of 20 meters per second given by the equation: h(t)=20t-5tsquared. When will the baton be back at the twirler's level? Aft

Question 142581: A twirler throws a baton with an initial upward velocity of 20 meters per second given by the equation: h(t)=20t-5tsquared. When will the baton be back at the twirler's level? After how many seconds will the baton reach its maximum height? :)
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
0=20t-5t^2 __ factoring __ 0=t(20-5t)

t=0 __ this is the starting position

20-5t=0 __ 20=5t __ 4=t __ this is the return to the start

time up=time down __ so max height is at 4/2 or 2 sec
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