A ball is thrown into the air with an upward

velocity of 40 ft/s. Its height h in feet after t seconds
is given by the function h = -16h 2 + 40t + 6.

A. In how many seconds does the ball reach its
maximum height?
B. What is the ball's maximum height?

A. V = Vo + g*t = 0,

40 + (-32)t = 0,
t = 1.25 s.

B . h = -16*1.25^2 + 40*1.25 + 6 = 31 Ft.

To find the maximum height of the ball and the time it takes to reach it, we need to determine the vertex of the parabolic function. The vertex of a quadratic function in the form h = at^2 + bt + c can be found using the formula t = -b / (2a).

Given that h = -16t^2 + 40t + 6, we can identify a = -16, b = 40, and c = 6.

A. In how many seconds does the ball reach its maximum height?

To find the time it takes for the ball to reach its maximum height, we will use the formula t = -b / (2a).

In this case, a = -16 and b = 40. Plugging these values into the formula, we get:

t = -40 / (2 * -16)
t = -40 / -32
t = 1.25 seconds

Therefore, the ball reaches its maximum height after 1.25 seconds.

B. What is the ball's maximum height?

To find the maximum height, we need to substitute the value of t into the function h = -16t^2 + 40t + 6.

Substituting t = 1.25 into the function, we get:

h = -16(1.25)^2 + 40(1.25) + 6
h = -16(1.5625) + 50 + 6
h = -25 + 50 + 6
h = 31 feet

So, the maximum height reached by the ball is 31 feet.

To find the answers to these questions, we need to analyze the given function and understand its properties. In this case, the function given is:

h = -16h^2 + 40t + 6

A. To determine when the ball reaches its maximum height, we need to find the vertex of the parabolic function h = -16h^2 + 40t + 6. The vertex is the highest point of the parabola, which represents the maximum height of the ball.

The formula for the x-coordinate of the vertex of a quadratic function in the form of h = ax^2 + bx + c is given by x = -b / (2a).

In this case, a = -16 and b = 40. Substituting these values into the formula, we have:

x = -(40) / (2(-16))
x = -40 / (-32)
x = 5/4 or 1.25 seconds

Therefore, the ball reaches its maximum height in 1.25 seconds.

B. To find the ball's maximum height, we can substitute the value of t (1.25 seconds) back into the function h = -16h^2 + 40t + 6.

h = -16(1.25)^2 + 40(1.25) + 6

Simplifying the equation, we get:

h = -16(1.5625) + 50 + 6
h = -25 + 50 + 6
h = 31 feet

Therefore, the ball's maximum height is 31 feet.

If your expression mean h = - 16 t² + 40 t + 6 then:

Vertex of this quadratic function have coordinates:

t = - b / 2a , h = c - b² / 4a

In this case:

a = - 16 , b = 40 , c = 6

So:

t = - b / 2a = - 40 / [ 2 ∙ ( - 16 ) ] = - 40 / - 32 = ( - 8 ) ∙ 5 / ( - 8 ) ∙ 4 = 5 / 4 = 1.25 sec

h = hmax = c - b² / 4a = 6 - 40² / [ 4 ∙ ( - 16 ) ] = 6 - 1600 / - 64 = 6 + 25 = 31 ft