a rock is thrown into the air from a height of 4 feet. the height of the rock above the ground, in feet, t seconds after the rock is thrown is given by -16t^2+56t+4. for how many seconds will the height of the rock be at least 28 feet above the ground

there will be a short time on the way up to 28 and a long one for the way down, we want the second minus the first

when will h = 28 feet?
28 = 4 + 56 t -16 t^2
16 t^2 - 56 t + 24 = 0
2 t^2 - 7 t + 3 = 0
(2t-1)(t-3) = 0
t = .5 or t = 3
so t = 2.5 seconds above 28

Well, isn't that a "rocking" question! To find out how many seconds the height of the rock will be at least 28 feet above the ground, we need to set up an inequality.

Let's set the height equation -16t^2 + 56t + 4 greater than or equal to 28:

-16t^2 + 56t + 4 ≥ 28

Now, let's subtract 28 from both sides:

-16t^2 + 56t + 4 - 28 ≥ 0

Simplifying the equation gives us:

-16t^2 + 56t - 24 ≥ 0

Now, let's solve this quadratic inequality. But instead of calculating it manually, I think I'll call my "mathemagician" friend to help us out. 🎩🐇

*poof*

Mathemagician: "Fear not, my friend! I shall solve this equation using the magical powers of factoring."

Upon solving the equation, we find that the rock's height will be at least 28 feet above the ground for a range of certain values of t seconds. The solution turns out to be:

2 ≤ t ≤ 7/2

Therefore, the height of the rock will be at least 28 feet above the ground for a period ranging from 2 seconds to 3.5 seconds.">

Remember, though, my calculations are based on assuming that the Earth's gravity is constant and that nothing else affects the rock's trajectory, like birds coming along and giving it a push. Anyway, I hope that helps!

To find out for how many seconds the height of the rock will be at least 28 feet above the ground, we need to solve the equation -16t^2 + 56t + 4 = 28.

First, let's rewrite the equation as a quadratic equation by subtracting 28 from both sides:

-16t^2 + 56t + 4 - 28 = 0

Simplifying further:

-16t^2 + 56t - 24 = 0

Next, let's divide the entire equation by -8 to simplify it:

2t^2 - 7t + 3 = 0

Now, we can factor the quadratic equation:

(2t - 1)(t - 3) = 0

Setting each factor to zero, we have:

2t - 1 = 0 or t - 3 = 0

Solving for t in each equation:

For 2t - 1 = 0:
2t = 1
t = 1/2

For t - 3 = 0:
t = 3

Therefore, the height of the rock will be at least 28 feet above the ground for t = 1/2 seconds and t = 3 seconds.

To determine the number of seconds the height of the rock will be at least 28 feet above the ground, we need to find the values of t that satisfy the inequality -16t^2 + 56t + 4 ≥ 28.

Step 1: Rewrite the inequality in standard form:
-16t^2 + 56t + 4 - 28 ≥ 0
-16t^2 + 56t - 24 ≥ 0

Step 2: Simplify the inequality:
-16t^2 + 56t - 24 ≥ 0

Step 3: Solve the inequality by factoring:
-4(4t^2 - 14t + 6) ≥ 0

Step 4: Factor the quadratic expression inside the parentheses:
-4(t - 2)(2t - 3) ≥ 0

Step 5: Determine the intervals where the inequality is true:

When the factor (t - 2) is greater than or equal to zero:
t - 2 ≥ 0
t ≥ 2

When the factor (2t - 3) is greater than or equal to zero:
2t - 3 ≥ 0
2t ≥ 3
t ≥ 3/2

So, the inequality is true for t ≥ 2 and t ≥ 3/2.

Step 6: Determine the intersection of the intervals:
Since the inequality is true for both t ≥ 2 and t ≥ 3/2, the intersection of these intervals is t ≥ 2.

Therefore, the height of the rock will be at least 28 feet above the ground for t ≥ 2 seconds.