The height is modelled by the equation:
[tex]h(t)=-0.2t^2+2t[/tex]
Therefore
[tex]\begin{gathered} \text{ } \\ \frac{\text{ dh}}{\text{ dt}}=-0.4t+2 \end{gathered}[/tex]
At the maximum height dh/dt = 0:
Hence,
[tex]\begin{gathered} -0.4t+2=0 \\ -0.4t=-2 \\ \text{ Dividing both sides by -0.4} \end{gathered}[/tex][tex]\begin{gathered} \frac{-0.4t}{-0.4}=\frac{-2}{-0.4} \\ t=5 \end{gathered}[/tex]
Hence, the ball gets to the maximum height after 5s.
The maximum height is given by h(5):
[tex]h(5)=-0.2(5)^2+2(5)=5[/tex]
Therefore, the maximum height is 5 ft.
When the ball reaches the ground h(t) = 0:
[tex]\begin{gathered} -0.2t^2+2t=0 \\ \text{ Dividing both sides by -0.2, we have:} \\ t^2-10t=0 \\ Factorising\text{ the left hand side, we have} \\ t(t-10)=0 \\ \text{ Therefore,} \\ t=0,\text{ t=10} \end{gathered}[/tex]
The ball reaches the ground in 10s
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