Exercise 9 Page 255

We are given a formula expressing the position of the baseball above the ground. h(t)=v_0t-16t^2 Let's substitute the given initial velocity, v_0=40, into the equation. h(t)=40t-16t^2The following graph illustrates the height of the ball above the ground between being hit and the time it comes back to the ground. Note that ignoring the height of the player means that we assume the ball starts at ground level.

To find the time it takes the ball to come back to ground, we need to solve h(t)=0. We can do this if we factor out t in the expression for h(t). 40t-16t^2=0 ⇕ t(40-16t)=0 Recall that a product is 0 if one of the factors is 0. t(40-16t)=0 ⇕ t=0 or 40-16t=0 The value t=0 corresponds to the time the ball is hit. To find the time when the ball is back on the ground, we need to find the solution of 40-16t=0.

It takes 2.5 seconds for the ball to fall back to the ground.