SOLUTION: 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 abouve the ground. Detrmine the m

Question 214441: 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 abouve the ground. Detrmine the maximum height of the ball to the nearest tenth of a meter.
Found 2 solutions by nerdybill, Earlsdon:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
h(t) = -4.9t^2+6t+0.6
.
Because of the negative coefficient associated with the t^2, we know that it is a parabola cup down. This means that if we find the "vertex" we will have the maximum.
.
The "axis of symmetry" is at:
-b/(2a)
= -6/(2*(-4.9))
= -6/(2*(-4.9))
= 0.612 seconds
This is the time it takes to reach the peak.
.
To find the height, plug the above back into:
h(t) = -4.9t^2+6t+0.6
h(0.612) = -4.9(0.612)^2+6(0.612)+0.6
h(0.612) = 2.4 meters

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Find the maximum height:

First find the time t, at which the ball reaches its maximum height.
This can be found by noting that the curve described by this quadratic equation (this is necessarily the trajectory of the ball) is a parabola that open downward. The maxumum point on this parabola occurs at the vertex of the parabola.
The x-coordinate (t-coordinate in this problem) is found by:
where, from the given equation, a = -4.9 and b = 6.

seconds. Now substitute this value of t into the given equation and solve for , the maximum height.

meters.
The maximum height of the ball is 2.4 meters.

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