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?

Step-by-step explanation:

Height:  h(t)= -0.2t^2+2t  =  2t(-0.1t + 1)

-0.2t^2+2t  is  a quadratic expression with a = -0.2 and b = 2.

The ball reaches its max height at t = -b/(2a).

Here, t = -2 / (2·[-0.2] ), or  t = 2/0.4 = 5 (sec)

The ball reaches its max height at at 5 sec.

The max ht. of the ball is h(5) = -0.2(5 sec)^2 + 2(5 sec) = (-5 + 10) sec, or 5 ft.

To determine after how many sec the ball reaches the ground, we set h(t) = 0 and solve for t:  

h(t)= -0.2t^2+2t = 0 = 2t(-0.1t + 1).  Thus, t = 0 sec or t = 10 sec

The ball reaches the ground again after 10 sec.

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.