Question

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.

Answer 1

Answer:

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.