A ball is thrown directly upward from a height of 4 ft with an initial velocity of 24 ?ft/sec

To find the maximum height we need to calculate the differentiation of given function s(t) =  -16t²+ 24t + 4

s'(t) = - 2 * 16 * t + 24 ( differentiation of tⁿ = nt^n-1 )

s'(t) = -32t + 24

now we do the first order derivation equal to 0;

-32t + 24 = 0

t = 24/32

t = 3/4;

now we check the second order derivative i.e.;

s''(t) = -32;

which is a constant and it is second order derivative ( even ) than there will be a maximum value at t = 3/4;

so at t = 3/4 there will be maximum height

and maximum height at t = 3/4 will be;

s(t) =  -16t²+ 24t + 4

s(3/4) =  -16(3/4)²+ 24(3/4) + 4

s(3/4) = -9 + 18 + 4

s(3/4) = 13 ft.