To determine the initial height, we can look at the equation h(t) = -4.9t^2 + 12t + 3. The initial height refers to the height of the ball when t = 0.
Plugging t = 0 into the equation, we get: h(0) = -4.9(0)^2 + 12(0) + 3 h(0) = 0 + 0 + 3 h(0) = 3
Therefore, the initial height of the ball is 3 meters.
To find the maximum height, we can look for the vertex of the quadratic equation. The vertex occurs at the highest point of the parabolic trajectory, where the ball reaches its maximum height.
The formula for the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b / (2a). In our case, a = -4.9 and b = 12.
x = -12 / (2 * -4.9) x ≈ 1.22
To find the corresponding y-coordinate (maximum height), we substitute this value back into the equation: h(1.22) = -4.9(1.22)^2 + 12(1.22) + 3 h(1.22) ≈ 11.6
Therefore, the maximum height of the ball is approximately 11.6 meters.
To determine when the ball hits the ground, we set h(t) equal to 0 and solve for t. This represents the time when the ball's height is zero, indicating it has returned to the ground.
-4.9t^2 + 12t + 3 = 0
We can solve this quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula in this case:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c: t = (-(12) ± √((12)^2 - 4(-4.9)(3))) / (2(-4.9))
Simplifying this equation will give us the solutions for t, which represent the times when the ball hits the ground.
To determine how long the ball is above 10 meters, we need to find the values of t for which h(t) is greater than 10.
-4.9t^2 + 12t + 3 > 10
Simplifying the equation will give us the range of t values for which the ball is above 10 meters.
Please note that I will calculate these values for you.