A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

a. Determine a function for the height of the projectile t seconds after it’s released.
b. How long does it take the projectile to reach of height of 100 feet on its way up?
c. How long is the projectile in the air?

height(t) = -16t^2 + 860t + 7

b) solve 100 = -16t^2 + 860t + 7 and use the smaller of the two t values you find

c) set height(t) = 0 and solve

a. To determine a function for the height of the projectile, we need to consider the equation of motion for a falling object or projectile. The equation for the height (h) at time (t) is given by:

h(t) = -16t^2 + vt + h0

where:
- h(t) is the height of the projectile at time t
- t is the time in seconds
- v is the initial velocity of the projectile
- h0 is the initial height of the projectile above the ground

Given that the projectile is fired directly upward, its initial velocity is positive (860 feet per second), and its initial height above the ground is 7 feet, we can substitute these values into the equation:

h(t) = -16t^2 + 860t + 7

b. To find the time it takes for the projectile to reach a height of 100 feet on its way up, we set h(t) = 100 and solve for t:

-16t^2 + 860t + 7 = 100

Rearranging the equation:

16t^2 - 860t - 93 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = -860, and c = -93. Plugging these values into the quadratic formula and solving for t will give us the time it takes for the projectile to reach a height of 100 feet on its way up.

c. To find the total time the projectile is in the air, we need to consider the time it takes for the projectile to reach its highest point and then come back down to the ground. The time it takes for the projectile to reach its highest point is half of the total time the projectile is in the air.

To find the highest point, we need to consider the vertex of the quadratic function -16t^2 + 860t + 7. The x-coordinate of the vertex can be found using the formula:

t = -b / (2a)

In this case, a = -16 and b = 860, so we can plug these values into the vertex formula to find the time it takes for the projectile to reach its highest point.

Once we know the time it takes for the projectile to reach its highest point, we can double this value to find the total time the projectile is in the air.

a. To determine a function for the height of the projectile, we can use the equation of motion for vertical motion:

h(t) = h0 + v0t - (1/2)gt^2

where:
- h(t) is the height of the projectile at time t
- h0 is the initial height (7 feet above the ground)
- v0 is the initial velocity (860 feet per second)
- g is the acceleration due to gravity (32 feet per second squared, assuming this is on Earth)
- t is the time in seconds

Substituting the given values into the equation, we get:

h(t) = 7 + 860t - (1/2)(32)t^2

b. To find the time it takes for the projectile to reach a height of 100 feet on its way up, we can substitute h(t) = 100 into the function for height and solve for t:

100 = 7 + 860t - (1/2)(32)t^2

Simplifying the equation:

0 = (1/2)(32)t^2 - 860t + 93

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Solving for t, we get:

t ≈ 1.0604 seconds or t ≈ 26.5016 seconds

Since the projectile is fired directly upward, we are only interested in the positive solution, t ≈ 1.0604 seconds.

c. As for how long the projectile is in the air, we need to consider the time it takes to reach its peak height as well as the time it takes to return to the ground. Since the projectile was fired directly upward, the time it takes to reach the peak height is the same as the time it takes to return to the ground.

Using the formula for time of flight in vertical motion:

T = 2 * (v0 / g)

Substituting the given values:

T = 2 * (860 / 32)

T ≈ 53.75 seconds

Therefore, the projectile is in the air for approximately 53.75 seconds.