math helped needed please

Question

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=72t-16t^2. What is the maximum height that the ball will reach?

Do not round your answer

Answer 1

h(x) = 72t - 16t^2 h in feet - t in seconds

This is a formula that models the velocity of a thrown ball.

When the ball has reached the maximum height acceleration is 0.

The formula for the acceleration for this equation is the derivative of the original equation.

This equation is 72 - 32t.

Set the equation to 0.

0 = 72 - 32t t in seconds

32t = 72

t = 2.25 seconds.

Subtitute 2.25 into the original equation.

72(2.25) - 16(2.25^2) = h in feet

162 - 16(5.0625) = h in feet

162 - 81 = h in feet

81 feet = h

Answer 2

f(x) = ax² + bx + c

Maximum is at -b/2a

h(x) = -16t² + 72t

Maximum is at -72 / 2(-16), or 2.25

h(2.25) = ?