Question

What is the time taken by an object to reach its maximum height when thrown upwards with an initial velocity of 40 ft/s? The height of the object is given by the function h(t) = -16t^2 + 40t + 5. Provide the maximum height reached by the object.

Answer 1

To determine the time taken by the object to reach its maximum height, we can analyze the given height function:

\[
h(t) = -16t^2 + 40t + 5
\]

This is a quadratic function in the form of \( h(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 40 \)
- \( c = 5 \)

The maximum height for a quadratic function occurs at the vertex, and the time \( t \) at which this occurs can be found using the formula:

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

Substituting the values of \( a \) and \( b \):

\[
t = -\frac{40}{2 \times -16} = -\frac{40}{-32} = \frac{40}{32} = \frac{5}{4} \text{ seconds} = 1.25 \text{ seconds}
\]

Now, to find the maximum height reached by the object, we substitute \( t = 1.25 \) back into the height function \( h(t) \):

\[
h(1.25) = -16(1.25)^2 + 40(1.25) + 5
\]

Calculating \( 1.25^2 \):

\[
1.25^2 = 1.5625
\]

Now substituting this value into the equation:

\[
h(1.25) = -16(1.5625) + 40(1.25) + 5
\]
\[
= -25 + 50 + 5
\]
\[
= 30 \text{ feet}
\]

Thus, the time taken to reach maximum height is \( 1.25 \) seconds, and the maximum height reached by the object is \( 30 \) feet.

**Final Answers:**
- Time taken to reach maximum height: **1.25 seconds**
- Maximum height reached: **30 feet**