I am a parent trying help my son, who has trouble understanding and remembering how to work his math. I never had this math, so it is hard for me to explain it. If I see it worked out. Then I can explain it, (I Think) here is the problems. A ball is tossed into the air and it is modeled by the function h(t)=-16t^2+96t+8. This has a maximum height of? Also need to find the time it would return to the ground?
h(t) is just a parabola. The max height is reached at the vertex. For the parabola
ax^2 + bx + c
the vertex is at x = -b/2a.
So, for h(t), the vertex is at t = 3.
h(3) = 152
It hits the ground when h=0. To find that, just use the quadratic formula.
To find the maximum height of the ball and the time it takes to return to the ground, you can use the formula for the vertex of a quadratic function. In this case, the function is h(t) = -16t^2 + 96t + 8, where 'h' represents the height of the ball at time 't'.
The formula to find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by x = -b / (2a). For this problem, we need to find the time at which the ball reaches its maximum height, so we'll use the formula with the coefficient of t^2 being 'a', the coefficient of t being 'b', and the constant term being 'c'.
Step 1: Find the time at which the ball reaches its maximum height (the x-coordinate of the vertex):
Formula: t = -b / (2a)
In this case, a = -16 and b = 96, so we have:
t = -96 / (2*-16)
t = -96 / -32
t = 3
The ball reaches its maximum height at t = 3.
Step 2: Find the maximum height (the y-coordinate of the vertex):
To find the maximum height, substitute the value of t = 3 into the function h(t):
h(t) = -16t^2 + 96t + 8
h(3) = -16(3)^2 + 96(3) + 8
h(3) = -16(9) + 288 + 8
h(3) = -144 + 288 + 8
h(3) = 152
The maximum height of the ball is 152 units.
Step 3: Find the time the ball will return to the ground:
To find when the ball will return to the ground, we need to find the time when the height (h(t)) is equal to 0.
0 = -16t^2 + 96t + 8
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
Formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 96, and c = 8. Substituting these values into the quadratic formula:
t = (-96 ± √(96^2 - 4*(-16)*8)) / (2*(-16))
t = (-96 ± √(9216 + 512)) / (-32)
t = (-96 ± √(9728)) / (-32)
t = (-96 ± √(64 * 152)) / (-32)
t = (-96 ± 8√(19)) / (-32)
Since we are looking for the time it takes to return to the ground, we only consider the positive solution:
t = (-96 + 8√(19)) / (-32)
Simplifying the expression:
t = (-12 + √(19)) / 4
Therefore, the time it will take for the ball to return to the ground is approximately (-12 + √(19)) / 4.