Solution:
The equation is given below as
[tex]H=20+40T-16T^2[/tex]
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
[tex]\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}[/tex]
Using the quadratic formula below, we will find the value of T as
[tex]\begin{gathered} T=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=16,b=-40,c=10 \\ by\text{ substituting the values, we will have} \\ T=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ T=\frac{-(-40)\pm\sqrt{(-40)^2-4(16\times10)}}{2\times16} \\ T=\frac{40\pm\sqrt{1600-640}}{32} \\ T=\frac{40\pm\sqrt{960}}{32} \\ T=\frac{40\pm30.98}{32} \\ T=\frac{40+30.98}{32},T=\frac{40-30.98}{32} \\ T=2.2seconds,T=0.3seconds \end{gathered}[/tex]
Graphically,
Hence,
The time at which the height of the helicopter will be H=30 is at
[tex]\Rightarrow T=2.2seconds,T=0.3seconds[/tex]
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
[tex]\begin{gathered} H=20+40T-16T^2 \\ H=0 \\ 20+40T-16T^2=0 \\ 16T^2-40T-20=0 \end{gathered}[/tex]
To figure out the value of T, we will use the formula below
[tex]\begin{gathered} 16T^{2}-40T-20=0 \\ T=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ T=\frac{-(-40)\pm\sqrt{(-40)^2-4(16\times-20)}}{2\times16} \\ T=\frac{40\pm\sqrt{1600+1280}}{32} \\ T=\frac{40\pm\sqrt{2880}}{32} \\ T=\frac{40\pm53.67}{32} \\ T=\frac{40+53.67}{32},T=\frac{40-53.67}{32} \\ T=\frac{93.67}{32},T=-\frac{13.67}{32} \\ T=2.9seonds,T=-0.4seconds \end{gathered}[/tex]
Hence,
The time at which the remote control helicopter will hit the ground at H=0 will be
[tex]\Rightarrow T=2.9seconds[/tex]
