Answer:
The maximum height reached = 20 feet
The time it takes to reach this height = 1 second
Explanations:
Given the equation that modeled the height of the ball "t" seconds after Susan throws it is expressed as:
[tex]h(t)=-16t^2+32t+4[/tex]
At maximum height, the velocity of the ball is zero. Therefore the velocity function is given as:
[tex]h^{\prime}(t)=v(t)=-32t+32[/tex]
Substituting v(t) = 0 to get the required time, we will have:
[tex]\begin{gathered} -32t+32=0 \\ -32t=-32 \\ t=1\text{second} \end{gathered}[/tex]
Therefore it takes the ball 1 second to reach its maximum height.
Determine the maximum height reached by substituting t = 1 sec into the formula as shown:
[tex]\begin{gathered} h(t)=-16t^2+32t+4 \\ h(1)=-16(1)^2+32(1)+4 \\ h(1)=-16+32+4 \\ h(1)=16+4 \\ h(1)=20\text{feet} \end{gathered}[/tex]
Hence the maximum height reached by the ball is 20 feet.
