A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by the quadratic function h left parenthesis t right parenthesis equals negative 4.9 t squared plus 9.8 t plus 1h(t)=−4.9t2+9.8t+1.

How long does it take for the baseball to reach its maximumheight? What is the maximum height obtained by the baseball?

It takes second(s) for the baseball to reach its maximum height.

Time to get maximum height   1 sec

Maximum height   5.9 m

Step-by-step explanation:

The height of the ball is according to problem statement:

h(t) = -4.9*t² + 9.8*t + 1

Taking derivatives on both sides of the equation we get:

dh/dt  = V(y) =  - 9.8*t + 9.8

h (maximum) will occurs when  V(y) = 0

Then

V(y) = 0     ⇒   -9.8*t + 9.8  = 0

-9.8*t = - 9.8

t = 9.8/9.8

t = 1 sec

And to get the maximum height we plugg the value of t in the equation of the height

h(t)  = - 4.9*t² + 9.8*t  + 1

h(t)  = - 4.9*(1)² + 9.8*(1) + 1

h(t)  = - 4.9 + 9.8  + 1

h(t) = 5.9 m

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

b.

step-by-step explanation: