Answer :
SOLUTION
We are given the equations below:
[tex]\begin{gathered} x(t)=2.5t \\ y(t)=-16t^2+3 \end{gathered}[/tex]PART A
The marble will hit the ground when y(t)=0
[tex]\begin{gathered} y(t)=-16t^2+3 \\ 0=-16t^2+3 \\ 16t^2=3 \\ t^2=\frac{3}{16} \\ t=\sqrt[]{\frac{3}{16}} \\ t=\frac{\sqrt[]{3}}{4} \\ t=0.4330\text{ seconds} \end{gathered}[/tex]The marble will hit the ground after 0.4330 seconds
PART B
How far horizontally from the edge of the table will the marble first hit the ground can be gotten by substituting the value of t into x(t).
[tex]\begin{gathered} x(t)=2.5\times0.4330 \\ x(t)=1.0825 \end{gathered}[/tex]Therefore, the distance horizontally from the edge of the table the marble hits first is 1.0825feet
PART C
To express the parametric equations above the single rectangular equation, we eliminate t from both equations.
[tex]\begin{gathered} x=2.5t \\ y=16t^2+3 \\ \text{therefore,}t=\frac{x}{2.5} \\ we\text{ substitute in y} \\ y=16(\frac{x}{2.5})^2+3 \\ y=\frac{16x^2}{6.25}+3 \end{gathered}[/tex]Therefore, the single rectangular equation is
[tex]y=\frac{16x^2}{6.25}+3[/tex]