The height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula:

$h = -16t^2 + 136t,$

where $h$ is the height in feet and $t$ is the time in seconds. To find the time required for the object to return to its point of departure, we need to determine when the height $h$ is zero (i.e., when the object returns to the ground).

Step 1: Set the height equation to zero

Set the height equation to zero:

$-16t^2 + 136t = 0.$

Step 2: Factor the equation

Factor out $t$:

$t(-16t + 136) = 0.$

This equation gives two solutions:

1. $t = 0$ (the starting time)
2. $-16t + 136 = 0$

Step 3: Solve for $t$

Solve the second equation:

$-16t + 136 = 0.$

$16t = 136.$

$t = \frac{136}{16}.$

Let's calculate the time required for the object to return to its point of departure.The time required for the object to return to its point of departure is:

$t = 8.5 \, \text{seconds}.$

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Would you like more details or have any questions? Here are some related questions you might find interesting:

1. What is the maximum height the object reaches, and at what time?
2. How would the height formula change if the initial velocity were different?
3. How long is the object in the air before reaching half of its maximum height?
4. What is the velocity of the object when it returns to the ground?
5. How would air resistance affect the time for the object to return to its starting point?

Tip: Quadratic equations often model projectile motion, where the coefficient of the $t^2$ term represents the effect of gravity.