a ball is thrown into the air with an upward velocity of 32 ft per second. Its height h in feet after t seconds is given by the function h(t)=-16t^2 +32t+6. How long does it take the ball to reach its maximum height? What is the balls maximum height?

a- 22 ft 1.0 second
b- 22 ft 2.0 seconds
c- 44 ft 2.17 seconds
d- 11 ft 2.17 seconds

my answer is c or d

h(t)=16t(2-t) + 6

so note that at t=0, and t=2 h(t) is 6, the starting point.

So max t occurs then at t=1, halfway (it is a parabola).

max h=16(1)+6=22

so it would be 22 ft after 1 second

Thank you very much

at 68 trying to help grandson I'm stretching a bit to when I took this

yes

Sports. A baseball is thrown into the air with upward velocity of 30 ft / s. Its height h, in feet, after t seconds is given by the function h = -16t ^ + 30t + 6. How long will it take the ball to reach its maximum height? What is the ball's maximum height? What is the range of the function?

To find the time it takes for the ball to reach its maximum height, we need to determine the time when the ball is at its highest point. This is the vertex of the parabolic function h(t)=-16t^2 + 32t + 6.

The t-coordinate of the vertex can be found using the formula t = -b/2a, where a, b, and c are the coefficients of the quadratic equation. In this case, a = -16 and b = 32.

t = -32 / (2*(-16))
t = -32 / (-32)
t = 1.0

So, the ball reaches its maximum height after 1 second.

To calculate the maximum height, substitute the value of t into the equation h(t):

h(t) = -16(1^2) + 32(1) + 6
h(t) = -16 + 32 + 6
h(t) = 22

Therefore, the ball reaches a maximum height of 22 ft.

The correct answer is option a: 22 ft, 1.0 second.