An object is thrown or fired straight upwards at an initial speed of v_0 ft⁄s will reach height h feet after t seconds, where h and t are related to the formula

h=-16t^2+v_0 t
Suppose the object is fired straight upwards with an initial speed of 800ft⁄s, What is the initial velocity?
How does it change the equation h=-16t^2+v_0 t?
What is the initial position of the object?
When does the object fall back to the ground?
When does the object reach a height of 6400ft?
When does the object reach a height of 2mi?
How high is the highest point the ball reaches?
Suppose the object is dropped from a height of 288ft, what is v_0?
The equation becomes h=-16t^2+h_0 after (g) Why?
Write an equation which includes 288ft

thank you Steve for your help

To answer these questions, let's break them down step by step:

1. What is the initial velocity?
The initial velocity, denoted as v_0, is given as 800 ft/s.

2. How does it change the equation h=-16t^2+v_0 t?
By substituting v_0 = 800 ft/s into the equation, it becomes h = -16t^2 + 800t.

3. What is the initial position of the object?
The initial position, often referred to as the initial height or h₀, is not directly given in the problem. To determine the initial position of the object, we need additional information.

4. When does the object fall back to the ground?
The object will fall back to the ground when its height, h, becomes 0. Plugging this into the equation h = -16t^2 + 800t and solving for t, we find the time it takes for the object to reach the ground.

0 = -16t^2 + 800t

Using factoring or the quadratic formula, we can solve for t.

5. When does the object reach a height of 6400 ft?
To find the time it takes for the object to reach a height of 6400 ft, we can set h = 6400 in the equation h = -16t^2 + 800t and solve for t.

6400 = -16t^2 + 800t

Solving this equation will give us the time it takes to reach a height of 6400 ft.

6. When does the object reach a height of 2 mi?
To find the time it takes for the object to reach a height of 2 mi (which is equal to 2 * 5280 ft = 10560 ft), we can set h = 10560 in the equation h = -16t^2 + 800t and solve for t.

10560 = -16t^2 + 800t

Solving this equation will give us the time it takes to reach a height of 2 mi.

7. How high is the highest point the ball reaches?
The highest point the ball reaches (also known as the peak or vertex) is the maximum value of the height function h = -16t^2 + 800t. To find this value, we can determine the t-coordinate of the vertex using the formula t = -b/2a, where a = -16 and b = 800. Substituting these values into the formula will give us the time at which the ball reaches its highest point. We can then substitute this time back into the height equation to find the corresponding height.

8. Suppose the object is dropped from a height of 288 ft, what is v_0?
When the object is dropped (not thrown), the initial velocity, v_0, is 0 ft/s. This is because there is no initial upward speed when an object is dropped, only the gravitational acceleration downward.

9. The equation becomes h=-16t^2+h_0 after (g) Why?
When an object is dropped (not thrown), the initial velocity, v_0, is 0 ft/s. In the presence of gravity, the equation for the height becomes h = -16t^2 + h₀, where h₀ represents the initial position or height from which the object was dropped.

10. Write an equation which includes 288 ft.
If we drop an object from a height of 288 ft, the equation for the height becomes h = -16t^2 + 288. This equation accounts for the initial drop from a height of 288 ft and the effect of gravity (-16t^2).

To find the initial velocity, we will use the given formula h = -16t^2 + v_0t and input the known values: h = 0 (as the object starts from the ground) and v_0 = 800 ft/s.

0 = -16t^2 + 800t

Next, we can rearrange the equation to solve for t:

16t^2 - 800t = 0

By factoring out 16t:

16t(t - 50) = 0

This equation yields two possible solutions: t = 0 (which indicates the time when the object is initially launched) and t = 50 seconds (when the object reaches its highest point and starts its descent back to the ground).

Now let's find the initial position of the object. The initial position corresponds to the height when the object is launched, which is when t = 0 seconds. Plugging t = 0 into the equation h = -16t^2 + v_0t:

h = -16(0)^2 + 800(0)
h = 0

This confirms that the initial position of the object is at a height of 0 feet above the ground.

To determine when the object falls back to the ground, we need to find the time when h = 0. Plugging this into our original equation:

0 = -16t^2 + 800t

We can solve for t by factoring out a t:

t(-16t + 800) = 0

This gives us two possible solutions: t = 0 (the initial launch) and t = 50 seconds (when the object falls back to the ground).

Now, let's find the time when the object reaches a height of 6400 feet. We can set up the equation using h = 6400:

6400 = -16t^2 + 800t

Rearranging the equation:

16t^2 - 800t + 6400 = 0

To simplify the equation, we can divide all terms by 16:

t^2 - 50t + 400 = 0

This equation can be factored into:

(t - 40)(t - 10) = 0

We now have two possible solutions: t = 40 seconds and t = 10 seconds. However, t = 10 seconds would represent the initial launch, so our answer is t = 40 seconds.

Next, let's find the time when the object reaches a height of 2 miles. Since 1 mile is equal to 5280 feet, 2 miles is equal to 10560 feet. Plugging this into our equation:

10560 = -16t^2 + 800t

Simplifying and rearranging the equation:

16t^2 - 800t + 10560 = 0

Dividing by 16:

t^2 - 50t + 660 = 0

This equation cannot be easily factored, so we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Using a = 1, b = -50, and c = 660 in the quadratic formula, we find:

t ≈ 39.47 seconds or t ≈ 10.53 seconds

Considering the object is initially launched at t = 0 seconds, the time when the object reaches a height of 2 miles is approximately t = 39.47 seconds.

To find the highest point the object reaches, we can determine the vertex of the parabolic equation h = -16t^2 + 800t. The vertex of a parabola in the form y = ax^2 + bx + c is given by (-b/2a, f(-b/2a)).

For our equation, the vertex is (-800/(2*(-16)), f(-800/(2*(-16)))). Simplifying:

(-800/(-32), f(-800/(-32))) = (25, 5000)

Therefore, the highest point the object reaches is at a height of 5000 feet.

If the object is dropped from a height of 288 feet without any initial velocity, then v_0 = 0. Plugging this value into the equation h = -16t^2 + v_0t:

h = -16t^2 + 0t + 288
h = -16t^2 + 288

After dropping the object from a height h_0, the equation becomes h = -16t^2 + h_0. In this case, since the object is dropped from a height of 288 feet, the equation becomes:

h = -16t^2 + 288

I hope this explanation helps! Let me know if you have any further questions.

initial velocity is 800 ft/s upward

it does not change the equation.

initial position is 0, since it was fired from the ground (height=0)

It falls back to the ground when h=0. So, solve for t in 800t-16t^2 = 0

Solve for t in 800-16t^2 = 6400
(why are there two solutions?)

Same as above, but you need to convert 2 miles to feet.

max height at the vertex of the parabola, when t is midway between the roots of the equation.

if dropped, v_0 is zero.

with v_0 = 0, the term vanishes, and we have an initial height, rather than initial speed.

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