Answer:
A model for height h of the penny t seconds after the penny dropped is;
[tex]h=500+10t-16t^2[/tex]Explanation:
Given that the penny was dropped from a hot-air balloon that moves at a constant rate of;
[tex]u_0=10ft\text{/s}[/tex]Note that the penny will also have the same initial rate.
At the point when the penny was dropped they are already at height;
[tex]h_0=500ft[/tex]The gravitational pull on the penny (acceleration due to gravity);
[tex]a=-32ft/s^2[/tex]The height of the penny at time seconds after it was dropped can be modelled using the equation of motion;
[tex]h=h_0+u_0t+\frac{1}{2}at^2_{}_{}^{}[/tex]Where;
[tex]\begin{gathered} h_0=initial\text{ height} \\ u_0t=height\text{ covered by initial velocity} \\ \frac{1}{2}at^2=height\text{ covered by gravitational pull} \end{gathered}[/tex]Substituting the given values we have;
[tex]\begin{gathered} h=h_0+u_0t+\frac{1}{2}at^2_{} \\ h=500+10t+\frac{1}{2}(-32)t^2 \\ h=500+10t-\frac{32}{2}t^2 \\ h=500+10t-16t^2 \end{gathered}[/tex]Therefore, a model for height h of the penny t seconds after the penny dropped is;
[tex]h=500+10t-16t^2[/tex]