This is a simple inequality problem.

`h(t) = -16t^2+64t+4`

The requirement is `h(t)gt= 52` .

`-16t^2+64t+4>=52`

`16t^2-64t-4<=-52`

Muliplication by negative 1 changes the inequality.

`16t^2-64t+48<=0`

Dividing by 16,

`t^2-4t+3<=0`

`(t-3)(t-1)lt= 0`

Now let's see how we can find the correct range.

There are three ranges.

`tlt=1`    and `1lt=tlt=3`     and `tgt=3`

Now let's substitute values in those ranges respectively and check for the sign. (The correct range should give negative as in the inequality)

`tlt=1` , Let's substitute t = 0,

`(t-3)(t-1) = (0-3)(0-1) = +3`  positive

1<=t<=3, Let's substitute t = 2,

`(t-3)(t-1) = (2-3)(2-1) = -1`

This range give negative result.

t=>3, Let's substitute t = 4,

`(t-3)(t-1) = (4-3)(4-1) = +3`       Positive.

Therefore the range which satisfies the inequality is,

`1lt=tlt=3`

Therefore between t =1 to t=3, the height of the ball is greater than or equal to 52.

We want to know when `h(t)` , given as `h(t)=-16t^2+64t+4` , is greater than or equal to 52. We solve an equivalent inequality (equivalent in the sense that solutions to one inequality are solutions to the other inequality).

Note that the solution will be an interval.

Write algebraically:

`-16t^2+64t+4>=52`

`-16t^2+64t-48>=0`         Subtract 52 from each side

`-16(t^2-4t+3)>=0`           Factor out a common -16

`-16(t-3)(t-1)>=0`             Factor the trinomial

Now the expression on the left is zero at t=1 and t=3. The graph of the function is a parabola, opening down, with a vertex at `t=-b/(2a)=-64/(2(-16))=2` . Thus the expression on the left is positive for `1<t<3` .

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The solution to the problem is `1<=t<=3`

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The graph of `h(t)` and the line `h=52` :