Question

PLEASE PLEASE HELP AND I NEED YOU TO DO IT FAST BECAUSE I HAVE TO GO SOON

Answer 1

The height is modelled by the equation:

`h(t)=-0.2t^2+2t`

Therefore

`\begin{gathered} \text{ } \\ \frac{\text{ dh}}{\text{ dt}}=-0.4t+2 \end{gathered}`

At the maximum height dh/dt = 0:

Hence,

`\begin{gathered} -0.4t+2=0 \\ -0.4t=-2 \\ \text{ Dividing both sides by -0.4} \end{gathered}`
`\begin{gathered} (-0.4t)/(-0.4)=(-2)/(-0.4) \\ t=5 \end{gathered}`

Hence, the ball gets to the maximum height after 5s.

The maximum height is given by h(5):

`h(5)=-0.2(5)^2+2(5)=5`

Therefore, the maximum height is 5 ft.

When the ball reaches the ground h(t) = 0:

`\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}`

The ball reaches the ground in 10s

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