Question

The path of a rolling marble rolling off of a desk can be modeled by the parametric equations: x(t)= 2.5t y(t)=-16t^2+3, In these equations the horizontal and vertical positions of the marble x(t) and y(t) are in the feet and t is measured in seconds. A) The marble will hit the ground after how many seconds? B) Using your answer to A, how far horizontally from the edge of the table will the marble first hit the ground? C) The parametric equations above can be expressed as the following single rectangular equation:

Answer 1

We are given the equations below:

`\begin{gathered} x(t)=2.5t \\ y(t)=-16t^2+3 \end{gathered}`

PART A

The marble will hit the ground when y(t)=0

`\begin{gathered} y(t)=-16t^2+3 \\ 0=-16t^2+3 \\ 16t^2=3 \\ t^2=(3)/(16) \\ t=\sqrt[]{(3)/(16)} \\ t=\frac{\sqrt[]{3}}{4} \\ t=0.4330\text{ seconds} \end{gathered}`

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).

`\begin{gathered} x(t)=2.5*0.4330 \\ x(t)=1.0825 \end{gathered}`

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.

`\begin{gathered} x=2.5t \\ y=16t^2+3 \\ \text{therefore,}t=(x)/(2.5) \\ we\text{ substitute in y} \\ y=16((x)/(2.5))^2+3 \\ y=(16x^2)/(6.25)+3 \end{gathered}`

Therefore, the single rectangular equation is

`y=(16x^2)/(6.25)+3`