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

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$