Question

Write a quadratic formula to model the vertical motion for each situation, give h(t)=-16^2+vsubscript0t+hsubscript0. Find the maximum height.

Initial vertical velocity: 128ft/s; initial height: 82 ft

Answer 1

To model the vertical motion given the initial conditions, we can use the formula for the height \( h(t) \) of an object in free fall:

\[
h(t) = -16t^2 + v_0 t + h_0
\]

where:
- \( v_0 \) is the initial vertical velocity (in feet per second),
- \( h_0 \) is the initial height (in feet),
- \( t \) is the time in seconds,
- The term \(-16t^2\) represents the effect of gravity on the object (assuming the acceleration due to gravity is approximately \( 32 \, \text{ft/s}^2\) downward).

Given:
- \( v_0 = 128 \, \text{ft/s} \)
- \( h_0 = 82 \, \text{ft} \)

We can substitute these values into the formula:

\[
h(t) = -16t^2 + 128t + 82
\]

To find the maximum height, we first identify the vertex of the parabola represented by this quadratic equation \( h(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 128 \)
- \( c = 82 \)

The time \( t \) at which the maximum height occurs can be found using the formula:

\[
t = -\frac{b}{2a}
\]

Substituting in our values:

\[
t = -\frac{128}{2 \times -16} = -\frac{128}{-32} = 4 \, \text{seconds}
\]

Now, we can calculate the maximum height \( h(t) \) at \( t = 4 \) seconds:

\[
h(4) = -16(4^2) + 128(4) + 82
\]
\[
= -16(16) + 128(4) + 82
\]
\[
= -256 + 512 + 82
\]
\[
= 256 + 82
\]
\[
= 338 \, \text{ft}
\]

Thus, the maximum height the object reaches is:

\[
\text{Maximum Height} = 338 \, \text{ft}
\]