Question

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?

Answer 1

A) The initial height is 32

B) The higher height that the ball gets is 68

C) at
`t\approx 3.56`

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.</p><p></p><p><strong>[tex]x_(1,2)=(-b\pm √(b^2-4ac))/(2a) = (-(48) \pm √(b(48)^2-4(-16)(32)))/(2(-16)) = (-48\pm √(2304+2048))/(-32)\\\\=(-48+√(4352))/(-32)\approx 3.56`