**Solution: We know that \(h=-5\left(t - 20\right)^2+180\).**

We will first find the value for ‘t’ for which ‘h’ will be maximum.

For ‘h’ to be maximum, \(-5\left(t-20\right)^2\) should be maximum. Since \(-5\left(t-20\right)^2\) is a perfect square, therefore, \(-5\left(t-20\right)^2\) ≥ 0.

But, \(-5\left(t-20\right)^2\) will be ≤ 0 [By the property of reverse inequality]

So, for ‘h’ to be maximum \(-5\left(t-20\right)^2\) = 0

=> \(-5\left(t-20\right)^2\)

=> \(\left(t-20\right)^2\)

=> (t − 20) = 0

=> t = 20.

‘5’ seconds after ball has reached maximum height ‘h’ at t = 20 + 5 = 25.

=> \(h=-5\left(25-20\right)^2+180\)

=> \(h=-5\left(5\right)^2+180\)

=> h = −5 * 25 + 180

=> h = -125 + 180

=> h = 55 feet

**A is the correct answer.**

Answer A