A ball thrown up in the air is at a height of h feet, t seconds after it was thrown,

Solution: We know that \(h\ =\ -3\left(\ t\ -\ 10\right)^2\ +\ 250\)

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

For ‘h’ to be maximum, \(-3\left(\ t\ -\ 10\right)^2\ \) should be maximum. Since \(\left(\ t\ -\ 10\right)^2\ \) is a perfect square, therefore, \(\left(\ t\ -\ 10\right)^2\ \) ≥ 0.

But, \(\left(\ t\ -\ 10\right)^2\ \) will be ≤ 0 [By the property of reverse inequality]

So, for ‘h’ to be maximum \(-3\left(\ t\ -\ 10\right)^2\ \)= 0

=> \(-3\left(\ t\ -\ 10\right)^2\ \) = 0

=> \(\left(\ t\ -\ 10\right)^2\ \) = 0

=> (t − 10) = 0

=> t = 10.

‘7’ seconds after ball has reached maximum height ‘h’ at t = 10 + 7 = 17.

=> \(h\ =\ -3\left(\ t\ -\ 10\right)^2\ +\ 250\)

=> \(h\ =\ -3\left(\ 17\ -\ 10\right)^2\ +\ 250\)

=> h = −3 * 49 + 250

=> h = -147 + 250

=> h = 103 feet

Answer B