Question

A baseball is thrown in a parabolic arc. It's position above the

ground at a given point in time can be represented by the quadratic
function p (t) = gt² + vot + po, where t≥ 0, g is -32 ft/sec/sec,
vo is initial velocity, and po is its initial position above the ground. If
the ball was thrown straight up at 24 ft/sec when it was 5 ft above
the ground, how high did it go? (Type the number, do not include
units in your answer)

Answer 1

The ball reached a maximum height of 9.5 feet above the ground.

How did we get the value?

To find the maximum height reached by the baseball, determine the vertex of the parabolic function
`\(p(t) = gt^2 + v_0t + p_0\),` where:

-
`\(g = -32\)` ft/sec² (acceleration due to gravity),

-
`\(v_0 = 24\)` ft/sec (initial velocity), and

-
`\(p_0\)` is the initial position above the ground.

The formula for the
`\(t\)`-coordinate of the vertex of a quadratic function
`\(ax^2 + bx + c\)` is given by
`\(t = -(b)/(2a)\)`. In our case,
`\(a = g\)`,
`\(b = v_0\)`, and
`\(c = p_0\)`.

So, the
`\(t\)`-coordinate of the vertex is
`\(t = -(v_0)/(2g)\)`.

Substitute the given values:

`\[ t = -(24)/(2 \cdot (-32)) \]`

`\[ t = (24)/(64) \]`

`\[ t = (3)/(8) \]`

Find the corresponding
`\(p(t)\)` value for this
`\(t\)`-coordinate to determine the maximum height.

`\[ p\left((3)/(8)\right) = g\left((3)/(8)\right)^2 + v_0\left((3)/(8)\right) + p_0 \]`

`\[ p\left((3)/(8)\right) = -32 \cdot (9)/(64) + 24 \cdot (3)/(8) + p_0 \]`

`\[ p\left((3)/(8)\right) = -(9)/(2) + 9 + p_0 \]`

`\[ p\left((3)/(8)\right) = -(9)/(2) + (18)/(2) + p_0 \]`

`\[ p\left((3)/(8)\right) = (9)/(2) + p_0 \]`

The expression
`\((9)/(2) + p_0\)` represents the maximum height above the ground. Since the ball was thrown straight up when it was 5 ft above the ground,
`\(p_0 = 5\)`.

So, the maximum height is
`\( (9)/(2) + 5 = (19)/(2) \)`, which simplifies to
`\( (19)/(2) = 9.5 \)`.

Therefore, the ball reached a maximum height of 9.5 feet above the ground.