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A ball is kicked into the air and follows the path described by h(t) = -4.9t2 + 6t + 0.6, where t is the time in seconds, and h is the height in meters above the ground. Find the maximum height of the ball. What value would you have to change in the equation if the maximum height of the ball is more than 2.4 meters?

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To find the maximum height of the ball, we need to determine the vertex of the quadratic equation. The vertex of a quadratic equation in the form h(t) = at^2 + bt + c is given by the formula t = -b / (2a).

In this case, a = -4.9, b = 6, and c = 0.6.

Substituting these values into the formula, we have:

t = -6 / (2 * (-4.9))

t = -6 / (-9.8)

t = 0.612

The maximum height occurs at t = 0.612 seconds.

To find the maximum height, substitute this value back into the equation:

h(0.612) = -4.9(0.612)^2 + 6(0.612) + 0.6

h(0.612) ≈ 1.856 meters

The maximum height of the ball is approximately 1.856 meters.

If the maximum height of the ball needs to be more than 2.4 meters, we would have to change the value of the constant term in the equation (the "c" value) to a value greater than 2.4.

User Bowen Peng
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