A projectile is launched upward from the ground at an initial velocity of 400 feet per second. During it’s upward flight, the projectile passes a hot air balloon floating at a constant height of 2400 feet. How many seconds later will the projectile pass the hot air balloon again on the way down?

you want to solve for t in

400t - 16t^2 = 2400

Then find the difference in the two times.

To determine the time it takes for the projectile to pass the hot air balloon again on the way down, we need to find the time it takes for the projectile to reach its maximum height and then come back down to the height of the hot air balloon.

First, let's find the time it takes for the projectile to reach its maximum height. We can use the equation of motion:

h = v_i * t - (1/2) * g * t^2

where:
h is the height of the projectile (in this case, the maximum height)
v_i is the initial velocity (400 feet per second)
g is the acceleration due to gravity (32 feet per second squared)
t is the time it takes for the projectile to reach the maximum height

At the maximum height, the velocity of the projectile will be zero. So we can set v_i - g * t = 0 and solve for t:

400 - 32t = 0
32t = 400
t = 400/32
t ≈ 12.5 seconds

Now, let's find the time it takes for the projectile to come back down to the height of the hot air balloon. The height that the projectile needs to descend is the sum of the initial height (0 feet) and the height of the hot air balloon (2400 feet). We can use the equation of motion again:

h = v_i * t + (1/2) * g * t^2

where:
h is the height of descent (2400 feet)
v_i is the initial velocity (400 feet per second)
g is the acceleration due to gravity (32 feet per second squared)
t is the time it takes for the projectile to reach the height of the hot air balloon on the way down

2400 = 400t + (1/2) * 32 * t^2

This is a quadratic equation, which we can solve to find the value of t.

Simplifying the equation:
32t^2 + 400t - 2400 = 0

Solving the quadratic equation:
t = (-b ± √(b^2 - 4ac))/(2a)

Using the quadratic formula:
t = (-400 ± √(400^2 - 4 * 32 * -2400))/(2 * 32)

Calculating the values:
t ≈ (-400 ± 486.83)/(64)

We can discard the negative value of t since time cannot be negative. Therefore, t ≈ (486.83 - 400)/(64) ≈ 1.36 seconds.

So, the projectile will pass the hot air balloon again on the way down approximately 1.36 seconds later.