23 You throw a basketball upward from the top of a 20foot hi

23. You throw a basketball upward from the top of a 20-foot high building. Its initial velocity is 32 feet per second. If the height of the basketball at timetis given by h -16t2 +32u +20 when does the ball reach its maximum height and how high is it at that time? A. maximum height of 36 feet after 1 second B. maximum height of 1 foot after 36 seconds C. maximum height of 36 feet after 2.5 seconds D. maximum height of 0 feet after 2.5 seconds

Solution

h = -16t^2 + 32t + 20

The height is max/min when velocity is equal to 0

v = h\'= -32t + 32 = 0
32t = 32
t = 1

And in this case, since the parabola has a negative leading term,
this t = 1 would be a maximum case

When t = 1, we get h = -16 + 32+ 20 = 36

So, answer :
max height of 36 when t = 1

Option A

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h = -16t^2 + 80t + 96

v = h\'= -32t + 80 = 0

32t = 80

t = 2.5

So, either options A or B

And the height would be a maximum
because of the negative leading term

-16(2.5)^2 + 80(2.5) + 96

196

Max height of 196 after 2.5 sec

Option A