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Sladkaya [172]
3 years ago
10

A ball is thrown upward from the top of a building. The ball's height above the ground after T seconds is given by the function:

h(t) = -16t^2+48t+32.
A. What is the initial height (i.e. the height of the building)?
B. How high did the ball go?
C. When does the ball hit the ground?
Mathematics
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

A) The initial height is 32

B) The higher height that the ball gets is 68

C) at t\approx 3.56

Step-by-step explanation:

A) For this part you only have to evaluate the function in t=0, this is:

h(0)= -16(0)^2 + 48(0)+32\\ \hspace{8em} = -16*0 + 48*0 + 32 = 32

B) To obtain the higher height you have to find the maximum value that reaches the function h(t). For  this we can use the first derivative rule.

Then we have:

h'(t)=-32t+48

How we only have one extreme point we assume that this is a maximum, and this point is in the value t\[tex] such that [tex]h'(t)=0, this is t= 1.5. Then, evaluate the function h(t) in t= 1.5

we have

h(1.5) = -16(1.5)^2 + 48(1.5)+32\\ = -16(2.25)+48(1.5)+32\\ = -36+72+32 = 68

C) In this case we need to know when the function is 0. For this we can use the quadratic formula, with a=-16,\ b=48,\ c=32[\tex] and taking the positive solution.[tex]x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(48) \pm \sqrt{b(48)^2-4(-16)(32)}}{2(-16)} = \frac{-48\pm \sqrt{2304+2048}}{-32}\\\\=\frac{-48+\sqrt{4352}}{-32}\approx 3.56

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