The height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula h = −16t2 + 136t, where h is the height in feet and t is the time in seconds. Find the time required for the object to return to its point of departure.


The problem involves finding the time required for the object to return to its point of departure, meaning when the height \( h \) is equal to zero.

The height of the object is given by the formula:
\[
h(t) = -16t^2 + 136t
\]
We need to find the value of \( t \) when \( h(t) = 0 \).

Setting the equation equal to zero:
\[
0 = -16t^2 + 136t
\]

This is a quadratic equation. We can factor out the common term \( t \):
\[
0 = t(-16t + 136)
\]

This equation gives us two solutions:
1. \( t = 0 \) (This is the initial time when the object is launched.)
2. \( -16t + 136 = 0 \)

Solving for \( t \) in the second equation:
\[
-16t + 136 = 0
\]
\[
-16t = -136
\]
\[
t = \frac{-136}{-16} = 8.5 \text{ seconds}
\]

Thus, the object will return to its point of departure after \( \mathbf{8.5} \) seconds.

Would you like further details or have any questions? Here are some related questions:

1. How high does the object go before it starts descending?
2. What is the velocity of the object at the highest point of its trajectory?
3. How long does it take for the object to reach its maximum height?
4. What is the significance of the coefficient \(-16\) in the height equation?
5. How would the equation change if the initial velocity was different?

**Tip:** When solving quadratic equations, always check if the equation can be factored easily before using the quadratic formula.

Learn how to find the time required for an object tossed upward with an initial velocity of 136 feet per second to return to its point of departure using quadratic equations. Detailed step-by-step solution included.

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