A rocket is launched upward and then falls to the ground. The height of the rocket above the ground is h= -16t^2 - 400t where h is the height of the rocket in feet and t is the number of seconds of the flight. How many seconds will it take for the rocket to fall back to the ground?

I am sure that the equation should have been h= -16t^2 + 400t. The way you have it would produce a negative value of h for every of t value you pick, which makes no sense

h will be zero at the start and when it returns to the ground.
So, ...
-16t^2 + 400t = 0
-16t(t - 25) = 0

then t=0 (the start) or
t = 25 (when it hits the ground)

A rocket is launched upward and then falls to the ground. The height of the rocket above the ground is h= -16t^2 - 400t where h is the height of the rocket in feet and t is the number of seconds of the flight. How many seconds will it take for the rocket to fall back to the ground?

The expression relating the height of a rocket as a function of time is h = Vo(t) - gt^2/2 where h is the maximum height reached in time t, Vo is the initial upward velocity and g = the deceleration due to gravity, 32.2 ft./sec.^2.

Your expression should therefore be h = 400t - 16t^2. With the initial velocity being 400 ft./sec., the distance traveled to zero velocity on the way up is hu = 400t - 16t^2.

When falling back to earth, the distance traveled is hd = 0(t) + 16t^2.

Since the distance up equals the distance down, 400t - 16t^2 = 0(t) + 16t^2 or 32t^2 = 400t or 400 = 32t making t = 25 sec., the time up and the time down, for a total trip time of 25 sec.

To find the time it takes for the rocket to fall back to the ground, we need to set the height equation equal to zero and solve for t.

We have the equation h = -16t^2 - 400t.

Setting h = 0, we get:
0 = -16t^2 - 400t.

Rearranging the equation, we have:
16t^2 + 400t = 0.

Factoring out a common factor of 16t, we have:
16t(t + 25) = 0.

Now we can set each factor equal to zero and solve for t:
16t = 0: t = 0.

t + 25 = 0: t = -25.

Since time cannot be negative in this context, we can ignore the solution t = -25.

Therefore, the rocket will take 0 seconds to fall back to the ground.

To find the time it takes for the rocket to fall back to the ground, we need to determine when the height (h) of the rocket is equal to zero.

We are given the equation for the height of the rocket as h = -16t^2 - 400t, where t represents the time in seconds and h represents the height in feet.

To find when the rocket reaches the ground, we set h equal to zero and solve for t:

0 = -16t^2 - 400t

To solve this quadratic equation, we can use the quadratic formula. The quadratic formula is given by:

t = (-b ± √(b^2 - 4ac)) / (2a)

In our case, the quadratic equation is in the form of at^2 + bt + c = 0, with a = -16, b = -400, and c = 0.

Using the quadratic formula:

t = (-(-400) ± √((-400)^2 - 4(-16)(0))) / (2(-16))
= (400 ± √(160000 - 0)) / (-32)
= (400 ± √160000) / (-32)
= (400 ± 400) / (-32)

Let's compute both solutions:

t1 = (400 + 400) / (-32) = 800 / (-32) = -25
t2 = (400 - 400) / (-32) = 0

We have two solutions: t = -25 and t = 0. However, time cannot be negative, so we discard the negative value.

Thus, it will take 0 seconds for the rocket to fall back to the ground.