A ball is thrown into the air with an upward velocity of 26 feet per
second. Its height, h, in feet after t seconds is given by the function h(t) = –16t^2 + 26t + 5.
a. How long does it take the ball to reach its maximum height?
Round to the nearest hundredth.
b. What is the ball’s maximum height? Round to the nearest
hundredth.
I am stumped on this. If you can help, please show work
don't forget your Algebra I. h(t) is just a quadratic, and the vertex (maximum height) is at t = -b/2a = 26/32 = 13/16
Now use that to find h(13/16)
I have this h(13/16). I can not get the steps of the problem to solve and get the correct answer of 0.81 seconds. The maximum ball's height to the answer 15.56 feet. I need help please.
To find the maximum height reached by the ball, we need to find the vertex of the quadratic function. The vertex of a quadratic function in the form of h(t) = at^2 + bt + c is given by the formula:
t = -b / (2a)
Let's find the values of a and b in our case:
h(t) = -16t^2 + 26t + 5
Here, a = -16 and b = 26. Substitute these values into the formula:
t = -26 / (2(-16))
t = -26 / (-32)
t = 0.8125
So, it takes approximately 0.8125 seconds for the ball to reach its maximum height.
To find the maximum height, substitute the value of t into the function h(t):
h(0.8125) = -16(0.8125)^2 + 26(0.8125) + 5
h ≈ 15.83
Therefore, the ball reaches a maximum height of approximately 15.83 feet.
huh?? !!
13/16 = 0.8125
h(13/16) = -16(13/16)^2 + 26(13/16) + 5 = 249/16 = 15.5625
or,
h(.8125) = -16*.8125^2 + 26*.8125 + 5 = -10.5625+21.125+5 = 15.5625
Too bad you couldn't be bothered to show your work ...