Answer:
A)
[tex]\begin{array}{|c|c|c|}\cline{1-3} t & H(t) & g(t)\\\cline{1-3} 1 & 60 & 20.4\\\cline{1-3} 2 & 76 & 30.8\\\cline{1-3} 3 & 60 & 41.2\\\cline{1-3} 4 & 12 & 51.6\\\cline{1-3} \end{array}[/tex]
Between 3 and 4 seconds.
B) The baseballs will collide between 3 and 4 seconds.
Step-by-step explanation:
Given functions:
[tex]H(t)=-16t^2+64t+12[/tex]
[tex]g(t)=10+10.4t[/tex]
Part A
Substitute the values of t = 1, 2, 3 and 4 into the two functions:
[tex]H(1)=-16(1)^2+64(1)+12=60[/tex]
[tex]H(2)=-16(2)^2+64(2)+12=76[/tex]
[tex]H(3)=-16(3)^2+64(3)+12=60[/tex]
[tex]H(4)=-16(4)^2+64(4)+12=12[/tex]
[tex]g(1)=10+10.4(1)=20.4[/tex]
[tex]g(2)=10+10.4(2)=30.8[/tex]
[tex]g(3)=10+10.4(3)=41.2[/tex]
[tex]g(4)=10+10.4(4)=51.6[/tex]
Create a table with the found values:
[tex]\begin{array}{|c|c|c|}\cline{1-3} t & H(t) & g(t)\\\cline{1-3} 1 & 60 & 20.4\\\cline{1-3} 2 & 76 & 30.8\\\cline{1-3} 3 & 60 & 41.2\\\cline{1-3} 4 & 12 & 51.6\\\cline{1-3} \end{array}[/tex]
The solution to H(t) = g(t) is between 3 and 4 seconds as:
When t = 3, H(t) > g(t)
When t = 4, H(t) < g(t)
To prove this, equate the equations and solve for t:
[tex]\implies -16t^2+64t+12=10+10.4t[/tex]
[tex]\implies -16t^2+53.6t+2=0[/tex]
Using the Quadratic Formula to solve for t:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
[tex]\implies t=\dfrac{-53.6 \pm \sqrt{53.6^2-4(-16)(2)} }{2(-16)}[/tex]
[tex]\implies t=\dfrac{53.6 \pm \sqrt{3000.96}}{32}[/tex]
[tex]\implies t=3.39, -0.04\:\:(\sf 2\:d.p.)[/tex]
As time is positive, t = 3.39 s (which is between 3 and 4 seconds).
Part B
When the two baseballs are at the same height they will collide.
Therefore, the baseballs will collide between 3 and 4 seconds (when t = 3.39 s).
