Question

A softball player has hit a ball into the air. The height, y, of the ball x seconds after it is hit is given byy= -16.x2 + 78x +3.25.-What is the initial height of the softball? | Select ]-How long was the softball in the air? [Select]-What is the maximum height of the softball? | Select ]-How long did it take for the softball to reach its maximum height? Select]

Answer 1

The given function is

`y=-16x^2+78x+3.25`

To answer the questions, we draw the function. The image below shows the graph.

The initial height of the softball is 3.25 because that's the height when x = 0, let's prove it.

`\begin{gathered} y=-16(0)^2+78(0)+3.25 \\ y=0 \end{gathered}`

As you can observe in the image, the softball was 5 seconds in the air, the x-intercept shows the moment when the softball reaches the ground.

The maximum height is 98.313 because that's the point where the ball stops and goes down.

The softball took 2.5 seconds to reach the maximum height.