Question

So it’s a and b questionsA) use the graph to determine when the helicopter is at H=30B) set up and solve an equation using the Quadratic Formula to determine when the remote controlled helicopter is at ground (H=0). Please report your answer to the first decimal place. Show all work One question multiple steps.

Answer 1

The equation is given below as

`H=20+40T-16T^2`

Step 1:

To determine the time when the height is at H=30, we will substitute the value of H=30 in the equation above and solve for T

`\begin{gathered} H=20+40T-16T^(2) \\ 30=20+40T-16T^2 \\ collect\text{ similar terms, we will have} \\ 16T^2-40T+30-20=0 \\ 16T^2-40T+10=0 \end{gathered}`

Using the quadratic formula below, we will find the value of T as

`\begin{gathered} T=(-b\pm√(b^2-4ac))/(2a) \\ a=16,b=-40,c=10 \\ by\text{ substituting the values, we will have} \\ T=(-b\pm√(b^2-4ac))/(2a) \\ T=(-(-40)\pm√((-40)^2-4(16*10)))/(2*16) \\ T=(40\pm√(1600-640))/(32) \\ T=(40\pm√(960))/(32) \\ T=(40\pm30.98)/(32) \\ T=(40+30.98)/(32),T=(40-30.98)/(32) \\ T=2.2seconds,T=0.3seconds \end{gathered}`

Graphically,

Hence,

The time at which the height of the helicopter will be H=30 is at

`\Rightarrow T=2.2seconds,T=0.3seconds`

Part B:

To figure out the time at which the height will be H=0, we will substitute the value of H=0 in the equation below and solve for T

`\begin{gathered} H=20+40T-16T^2 \\ H=0 \\ 20+40T-16T^2=0 \\ 16T^2-40T-20=0 \end{gathered}`

To figure out the value of T, we will use the formula below

`\begin{gathered} 16T^(2)-40T-20=0 \\ T=(-b\pm√(b^2-4ac))/(2a) \\ T=(-(-40)\pm√((-40)^2-4(16*-20)))/(2*16) \\ T=(40\pm√(1600+1280))/(32) \\ T=(40\pm√(2880))/(32) \\ T=(40\pm53.67)/(32) \\ T=(40+53.67)/(32),T=(40-53.67)/(32) \\ T=(93.67)/(32),T=-(13.67)/(32) \\ T=2.9seonds,T=-0.4seconds \end{gathered}`

Hence,

The time at which the remote control helicopter will hit the ground at H=0 will be

`\Rightarrow T=2.9seconds`