Question

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 maximum​height? What is the maximum height obtained by the​ baseball?

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

Answer 1

Time to get maximum height 1 sec

Maximum height 5.9 m

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