An object is thrown straight up from a height of \(48\) ft and its velocity, in ft/s, is given by \(v(t)=-32t+64\), for \(t\) in time interval \(I\), where \(t\) is time in seconds.
- (a)
- What is the object’s maximum velocity? \[v_{max}=\answer {64}\,ft/s\]
- (b)
- Find the position of the object at any time \(t\), for \(t\) in \(I\)? You have to solve the Initial Value Problem:
\(s'(t)=-32t+64\)
\(s(0)=48\)
\[s(t)=\answer {-16} t^2+\answer {64} t+\answer {48} \,ft/s\] - (c)
- When will the object reach a height of \(0\)? Solve the equation
\(s(t)=0\)
\[t=\answer {2+\sqrt {7}}\,s\] - (d)
- What is the object’s maximum displacement from its starting position?
\[d=\answer {64}\,ft\]
- (e)
- When does the maximum displacement occur? \[t=\answer {2}\,s\]