suppose h(t) = -0.2t^2 + 2t models the height, in feet, of a ball that is kicked into the air where t is given as time, in seconds. After how many seconds does the ball reach its maximum height? What is the maximum height of the ball? After how many seconds does the ball reach the ground?
To find the time at which the ball reaches its maximum height, we need to determine the vertex of the parabola represented by the equation h(t) = -0.2t^2 + 2t.
The t-coordinate of the vertex of a parabola given by the equation y = ax^2 + bx + c is given by the formula t = -b/2a.
In this case, a = -0.2, b = 2, so t = -2/(2*(-0.2)) = -2/(-0.4) = 5.
Therefore, the ball reaches its maximum height after 5 seconds.
To find the maximum height, we can substitute this value of t back into the equation h(t):
h(5) = -0.2*(5)^2 + 2*(5) = -0.2*25 + 10 = -5 + 10 = 5 feet.
So, the maximum height of the ball is 5 feet.
To find when the ball reaches the ground, we need to find the time at which the height is 0. Setting h(t) = 0:
0 = -0.2t^2 + 2t
0 = t(-0.2t + 2)
This equation is zero either when t = 0 or when -0.2t + 2 = 0.
Solving -0.2t + 2 = 0 gives t = 2/0.2 = 10.
Therefore, the ball reaches the ground after 10 seconds.