Part A
By evaluating the derivative with respect to t of the height function h(t),
which is h’(t)=-2t+4,
we can find the turning point of the quadratic graph by solving h’(t)=0.
-2t+4=0
t=2
Therefore, we can know that when t=2, Sally’s potato attains its maximum height, which is
-(2^2)+4(2)+5
=9
Given that Sam’s potato has been 10 unit high, Sam’s potato went higher than Sally’s did.
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Part B
Note that the for all t where h(t)>0, the potato is still in the air.
So all we need to do is to solve h(t)>0.
-t^2 + 4t + 5 > 0
t^2 - 4t - 5 < 0
(t + 1)(t - 5) < 0
Therefore we get -1 < t < 5.
But t represents time so it can’t be negative.
So the overall solution is 0 < t < 5.
Therefore the total flying time of Sally’s potato is 5 units.
Because the flying time of Sam’s potato is 3.5 units,
Sally’s potato stayed in the air longer.
