the height h(t) in feet of an object thrown into the air with an initial upward velocity of 63 feet per second is given by the formula h(t)=-16t^2+63t+20 where t is the time in seconds. What's the height of the object after 2 seconds?
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Question 1.

f(t)=-5t²+20t+5=5(-t²+4t+1)=

5(-t²+4t-4+4+1)=

5(5-(t-2)²). This is maximum when t=2, and the maximum height is therefore 5×5=25m.

Question 2.

h(2)=-64+126+20=82ft.

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