Question

A ball is kicked upward with an initial velocity of 68 feet per second. The ball's height, h (in feet), from the ground is modeled by h = negative 16 t squared 68 t, where t is measured in seconds. What is the practical domain in this situation? a. 0 less-than-or-equal-to t less-than-or-equal-to 4.25 b. All real numbers c. 0 less-than-or-equal-to t less-than-or-equal-to 2.125 d. 0 less-than-or-equal-to t less-than-or-equal-to 17

Answer 1

Answer: a. 0 ≤ t ≤ 4.25

Step-by-step explanation: To determine the practical domain in this situation, we need to consider the physical constraints of the problem. The practical domain refers to the range of values for the independent variable, t, that makes sense in the given context.

In this case, since we are modeling the height of a ball kicked upward, time (t) cannot be negative because it represents the duration since the ball was kicked. Therefore, the value of t must be non-negative.

Additionally, to find the time it takes for the ball to reach its maximum height and fall back to the ground, we can set the equation h = 0 and solve for t.

Using the given equation: h = -16t^2 + 68t

0 = -16t^2 + 68t

Dividing the equation by 4 gives us:

0 = -4t^2 + 17t

Factoring out t, we get:

0 = t(-4t + 17)

From this equation, we can see that one solution is t = 0, which represents the starting point when the ball is kicked.

The other solution is obtained when -4t + 17 = 0:

4t = 17

t = 17/4

t = 4.25

Therefore, the ball reaches the ground again at t = 4.25 seconds.

Considering the physical context, we can conclude that the practical domain for this situation is:

0 ≤ t ≤ 4.25

This corresponds to option (a) 0 ≤ t ≤ 4.25.