`h = 20t-5t^2`

Thus is related to maxima and minima in derivatives.

For maximum h then `(dh)/dt = 0`

`h = 20t-5t^2`

`(dh)/dt = 20-10t`

When `(dh)/dt = 0`

`20-10t = 0`

t = 2

If h is a maximum at `t=2` then `(d^2h)/dt^2 <...

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`h = 20t-5t^2`

Thus is related to maxima and minima in derivatives.

For maximum h then `(dh)/dt = 0`

`h = 20t-5t^2`

`(dh)/dt = 20-10t`

When `(dh)/dt = 0`

`20-10t = 0`

t = 2

If h is a maximum at `t=2` then `(d^2h)/dt^2 < 0` at t=2

`(dh)/dt = 20-10t`

(d^2h)/dt^2 = -10 <0

So h has a maximum at t=2

Maximum height above cliff `= 20*2-5*2^2 = 20m`

*So the maximum height above cliff is 20m.*

Since the editors can answer only one question i will left the other two for you. But you can use the following guides.

- When stone hits the beach what is h. h is (+) in upwards.you should remember that here.
- The derivative of h function will give you the velocity function. Then its a matter of substituting the time.