help me pleaseee!!!

Question

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t^2 + 64t + 4, where t is measured in seconds and h is the height in feet. Use this information to answer the following questions .

what is the maximum height of the ball?

Answer 1

Hi Christina B.

A review and simple comparison of the given function

h(t) = -16t² + 64t + 4

Should confirm that you have the following type of function

f(x) = ax² + bx + c; A Standard Form Quadratic or parabola

a = -16, b = 64 and c = 4

t is your x coordinate

h(t) is your y coordinate

You can use the properties associated with quadratics

First the coefficient of x in your function is -16 which means that your parabola opens downward

Second the vertex is a the maximum

Third the x coordinate of the vertex is (-b/2a) for your function

-b/2a = -64/2(-16) = -64/-32 = 2; so t = 2 seconds

You can use the x coordinate of the vertex 2 to find the y coordinate by simply plugging it into your original function

h(2) = -16(2²) + 64(2) + 4 = -16(4) + 128 + 4 = -64 + 128 + 4 = 68 feet

You can graph your function at Desmos.com to confirm this maximum.

Answer 2

68 feet

take the derivative, set equal to zero, solve for t, then plug that value into the original h(t) equation

h'(t) = -32t +64 = 0, t= 64/32 = 2 seconds

h(2) = -16(2)^2 + 64(2) + 4 = -64+128+4 = 68 feet = max height