A ball is thrown into the air with an upward velocity of 48 /. Its height ℎ in feet after seconds is given by the function ℎ = −162 + 48 + 9. How long does it take the ball to reach is the maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary.
please help
V = Vo + g*T = 0
48 + (-32)T = 0
T = 1.5 s.
thank you so much
its 4
To find the maximum height and the time it takes for the ball to reach it, we need to analyze the given function `h(t) = -16t^2 + 48t + 9`.
1. Maximum height:
The maximum height occurs at the vertex of the parabolic function. To find the vertex, we use the formula `t = -b / (2a)`, where `a` and `b` are the coefficients of the function. In this case, `a = -16` and `b = 48`.
t = -48 / (2 * -16)
t = -48 / -32
t = 3/2
t = 1.5 seconds
To find the maximum height, substitute the value of `t` into the function:
h(1.5) = -16(1.5)^2 + 48(1.5) + 9
Simplifying the equation:
h(1.5) = -16(2.25) + 72 + 9
h(1.5) = -36 + 72 + 9
h(1.5) = 45 feet
Therefore, the ball reaches its maximum height of 45 feet.
2. Time to reach the maximum height:
As we found earlier, it takes 1.5 seconds for the ball to reach its maximum height.
In summary:
The ball reaches its maximum height of 45 feet after 1.5 seconds.
You need to proofread your post. Clearly you meant
h = −16t^2 + 48t + 9
it reaches its maximum height when v=0
v = 48-32t
Now you have t, so plug it in to find h