A ball 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?FIND THE VETEX!!!!

For this case we have the following equation:
 h (t) = -4.9t ^ 2 + 6t + 0.6
 Deriving we have:
 h '(t) = -9.8t + 6
 We equal zero and clear t:
 -9.8t + 6 = 0
 9.8t = 6
 t = 6 / 9.8
 t = 0.61
 We substitute the value of time in the equation of the height:
 h (0.61) = -4.9 * (0.61) ^ 2 + 6 * (0.61) +0.6
 h (0.61) = 2.44m
 For the height to be higher than 2.44m, two values can be changed:
 Initial speed: 6
 Initial height: 0.6
 Answer:
 The maximum height of the ball is:
 h (0.61) = 2.44m
 You would have to change in the equation if the maximum height of the ball is more than 2.4 meters:
 Initial speed: 6
 or
 Initial height: 0.6