A projectile is fired straight upward from ground level with an initial velocity of 320 feet per second. (Assume t = 0 seconds corresponds to the time the object is fired. Use 32 feet/second2 as acceleration due to gravity.)

(a) At what instant will it be back at ground level?
(b) When will the height exceed 1584 feet? (Enter your answer using interval notation. Enter your answer in terms of seconds.)

(a) time up = time down

... = 320 fps / 32 ft/s^2

(b) the velocity at a given height is the same upward and downward EXCEPT for the direction

the max height is 1600 ft, which is reached in 10 sec ... the velocity is zero at this point

1584 ft is one sec below the max
... 9s < t < 11s

height = -16t^2 + 320t + 0

at ground level, height = 0
16t^2 - 320t = 0
16t(t - 20) = 0
t = 0 , at the start
or
t = 20

b) you want
-16t^2 + 320t > 1584
16t^2 - 320t + 1584 < 0
t^2 - 20t + 99 < 0
(t-11)(t-9) < 0

9 < t < 11 , where t is in seconds

(a) To find the instant when the projectile is back at ground level, we can use the equation of motion:

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

Where:
h = height of the projectile
h0 = initial height (ground level)
v0 = initial velocity
g = acceleration due to gravity (32 ft/s^2)
t = time in seconds

Since the projectile is fired straight upward from ground level, the initial height (h0) is 0.

So the equation becomes:

h = v0t - (1/2)gt^2

We want to find when the height (h) is again 0. Therefore, we can set the equation equal to 0:

0 = v0t - (1/2)gt^2

Simplifying the equation, we have:

(1/2)gt^2 = v0t

Dividing both sides by t, we get:

(1/2)gt = v0

Further simplifying, we have:

t = (2v0) / g

Substituting the given values:
v0 = 320 ft/s
g = 32 ft/s^2

t = (2 * 320) / 32
t = 640 / 32
t = 20 seconds

Therefore, the projectile will be back at ground level after 20 seconds.

(b) To find when the height exceeds 1584 feet, we can use the same equation of motion:

h = v0t - (1/2)gt^2

We want to find when the height (h) is greater than 1584 feet. So we have:

1584 < v0t - (1/2)gt^2

Rearranging the equation, we have:

(1/2)gt^2 - v0t + 1584 < 0

We can solve this quadratic inequality using techniques like factoring or the quadratic formula. However, to simplify the calculation, we can use the vertex form of a quadratic equation:

f(t) = a(t - h)^2 + k

Where (h, k) gives the coordinates of the vertex.

In our case, a = (1/2)g, h = v0 / g, and k = 1584.

Plugging in the values:

f(t) = (1/2)(32)(t - (320/32))^2 + 1584
f(t) = 16(t - 10)^2 + 1584

To find when f(t) is less than 0, we set it equal to 0 and solve for t:

16(t - 10)^2 + 1584 = 0

Now we can find the value of t:

16(t - 10)^2 = -1584
(t - 10)^2 = -1584 / 16
(t - 10)^2 = -99

Since a perfect square cannot be negative, there are no real solutions to this equation. Hence, the height never exceeds 1584 feet.

Therefore, the answer in interval notation is: (No solution)

To solve these questions, we can use the equations of motion for projectiles. Let's start by finding the time it takes for the projectile to reach its highest point and then return to the ground.

(a) At its highest point, the vertical velocity of the projectile will become zero. We can use the equation:
v = u + at

Where:
v = final velocity (0 ft/s)
u = initial velocity (320 ft/s)
a = acceleration due to gravity (-32 ft/s^2)
t = time taken

Rearranging the equation, we get:
t = (v - u) / a

Substituting the values:
t = (0 - 320) / -32
t = 10 seconds

So, it will take 10 seconds for the projectile to reach its highest point. Therefore, it will take the same amount of time for it to return to the ground.

(b) To find when the height exceeds 1584 feet, we need to determine the time at which the projectile reaches that height. We'll use the equation for displacement:
s = ut + (1/2)at^2

Where:
s = displacement (height) - 1584 ft
u = initial velocity (320 ft/s)
a = acceleration due to gravity (-32 ft/s^2)
t = time taken

Rearranging the equation, we get:
t^2 + (2u/a)t - (2s/a) = 0

Substituting the values:
t^2 + (64)t - 99 = 0

To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Where:
a = 1
b = 64
c = -99

Solving for t using the quadratic formula:
t = (-64 ± √(64^2 - 4(1)(-99))) / (2(1))
t ≈ (-64 ± √(4096 + 396)) / 2
t ≈ (-64 ± √4492) / 2
t ≈ (-64 ± 67) / 2

Taking the positive value:
t ≈ 3.5 seconds

Therefore, the height will exceed 1584 feet after 3.5 seconds.

In interval notation, we would write the answer as [3.5, ∞), meaning the height will exceed 1584 feet from 3.5 seconds onwards.