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

You may want to check the equation to see if there is a typo. I expect it to read:

h=-16t²+v_0 t
since the object is thrown straight upwards, while gravity (downwards) is indicated negative.

If it is fired straight upwards at an initial speed of 800 ft/s, the initial velocity is +800 ft/s, i.e. upwards.

yes, you're right. the equation is h=-16t^2+v_0t

To answer your questions:

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

2. The equation h = -16t^2 - v_0t represents the relationship between the height (h) and time (t) for an object thrown upwards with an initial velocity of v_0 ft/s. The equation can be derived from the equation of motion by considering the acceleration due to gravity (-32 ft/s^2) and the initial velocity.

3. Since the object is fired straight upwards, its initial position is the height from which it is fired. This information is not provided, so the initial position cannot be determined with the given information.

4. The object falls back to the ground when its height (h) becomes zero. Using the equation h = -16t^2 - v_0t and setting h = 0, you can solve for the time (t) at which the object falls back to the ground.

5. To determine when the object reaches a height of 6400 ft, substitute h = 6400 ft in the equation h = -16t^2 - v_0t and solve for time (t).

6. To determine when the object reaches a height of 2 mi (1 mi = 5280 ft), substitute h = 2 * 5280 ft in the equation h = -16t^2 - v_0t and solve for time (t).

7. The highest point reached by the object is the vertex of the parabolic equation h = -16t^2 - v_0t. The height can be determined by finding the vertex. The x-coordinate of the vertex, which gives the time (t) at the highest point, can be found using the formula t = -v_0 / (2 * (-16)). Substitute this value in the equation to find the height (h) at the highest point.

8. If the object is dropped from a height of 288 ft, the initial velocity, denoted as v_0, can be determined by equating the initial velocity term v_0t to zero in the equation h = -16t^2 - v_0t. Solving for v_0 will give you the value.

9. The equation h = -16t^2 + h_0 is obtained after dropping the object from a height of 288 ft. The term h_0 represents the initial height of the object. In this case, it is 288 ft. The acceleration due to gravity (-16t^2) remains the same, but the initial velocity term is dropped since the object starts from rest.

10. To write an equation including 288 ft, you can use h = -16t^2 - v_0t + h_0. Substituting v_0 = 0 (since the object is dropped), and h_0 = 288 ft, the equation becomes h = -16t^2 + 288 ft.

To answer these questions, we need to understand the given equation and how to interpret it.

1. What is the initial velocity?
The initial velocity (v₀) of the object is given as 800 ft/s.

2. How does it change the equation h=-16t^2-v₀t?
By substituting v₀ = 800 ft/s into the equation, it becomes h = -16t^2 - 800t.

3. What is the initial position of the object?
In the given equation, h represents the height of the object. Since the object is fired straight upwards, the initial height (h₀) would be 0 ft. Therefore, the initial position of the object is at ground level.

4. When does the object fall back to the ground?
The object falls back to the ground when its height (h) reaches 0 ft. To find the time (t) at which this occurs, we set the equation h = -16t^2 - 800t equal to 0 and solve for t. We then use the quadratic formula or factoring to find the roots, which will give us the times when the object reaches the ground.

5. When does the object reach a height of 6400 ft?
To find the time (t) at which the object reaches a height of 6400 ft, we set the equation h = -16t^2 - 800t equal to 6400 and solve for t. Again, we can use the quadratic formula or factoring to find the roots.

6. When does the object reach a height of 2 mi?
To find the time (t) at which the object reaches a height of 2 miles (which is 2 * 5280 = 10560 ft), we set the equation h = -16t^2 - 800t equal to 10560 and solve for t using the same methods as before.

7. How high is the highest point the object reaches?
The highest point the object reaches corresponds to the vertex of the parabolic trajectory. The x-coordinate of the vertex (t-coordinate in this case) can be found using the formula t = -b / (2a), where a = -16 and b = -800 from the given equation. Substitute this value of t into the equation to find the corresponding height (h).

8. Suppose the object is dropped from a height of 288 ft, what is v₀?
If the object is dropped instead of being fired upwards, the initial velocity (v₀) is 0 ft/s. This is because there is no initial upward velocity when it is dropped.

9. The equation becomes h = -16t^2 + h₀ after (g) Why?
When an object is dropped from a height (h₀) instead of being fired upwards, the initial velocity (v₀) is 0 ft/s. Since the initial velocity term (v₀t) disappears, the equation becomes h = -16t^2 + h₀.

10. Write an equation which includes 288 ft.
If we want to include the initial height (h₀) of 288 ft in the equation, it would be h = -16t^2 + 288.