Given the formula:
[tex]h(t)=150+40t-5t^2[/tex]Let's solve for the following:
• (a). What is its height after 4 seconds?
To find the height after 4 seconds, substitute 4 for t and solve for h(4):
[tex]\begin{gathered} h(4)=150+40(4)-5(4)^2 \\ \\ h(4)=150+160-80 \\ \\ h(4)=230 \end{gathered}[/tex]The height after 4 seconds is 230 feet.
• (b). What is its velocity after 4 seconds?
To find the velocity, we have:
[tex]v(t)=\frac{dh}{dt}=\frac{d}{dt}(150+40t-5t^2)[/tex]Now find the derivative:
[tex]v(t)=40-10t[/tex]Substitute 4 for t and solve for v(4):
[tex]\begin{gathered} v(4)=40-10(4) \\ \\ v(4)=40-40 \\ \\ v(4)=0\text{ m/s} \end{gathered}[/tex]Therefore, after 4 seconds, the velocity is 0 m/s.
ANSWER:
a) 230 ft
b) 0
