Suppose h(t) = -0.2t² + 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?
_____ seconds
What is the maximum height of the ball?
_____feet
After how many seconds does the ball reach the ground?
______ seconds
To find the maximum height of the ball, we need to determine when the derivative of h(t) is zero.
h(t) = -0.2t² + 2t
Taking the derivative of h(t) with respect to t: h'(t) = -0.4t + 2
Setting h'(t) equal to zero and solving for t:
-0.4t + 2 = 0
-0.4t = -2
t = -2 / -0.4
t = 5
Therefore, the ball reaches its maximum height after 5 seconds.
Plugging t = 5 into the equation for h(t) to find the maximum height:
h(5) = -0.2(5)² + 2(5)
h(5) = -0.2(25) + 10
h(5) = -5 + 10
h(5) = 5
Thus, the maximum height of the ball is 5 feet.
To find when the ball reaches the ground, set h(t) equal to 0 and solve for t:
-0.2t² + 2t = 0
t(-0.2t + 2) = 0
t = 0 or -0.2t + 2 = 0
-0.2t = -2
t = -2 / -0.2
t = 10
Therefore, the ball reaches the ground after 10 seconds.
h(t) = -0.2t² + 2t
when does h = 0?
0.2 t^2 - 2 t = 0
t ( 0.2 t -2) = 0
so at t = 0 (of course) and at t = 2 / 0.2 = 10
yes agree