Aball is kicked into the air and follows the path described by h(t)= – 4.9t^2 + 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?

The maximum value can be determined by taking the derivative of the function.

(dh/dt) [h(t)] = h'(x) = -9.8t + 6

Set h'(x) = 0 to find the critical point

-9.8t + 6 = 0
-9.8t = -6
t = 6/9.8

Plug the time back into the function to find the height.

h(6/9.8) = -4.9(6/9.8)^2 + 6(6/9.8) + .6
= 2.4

And I don't understand your second question.