Posted by **Tom** on Monday, March 3, 2008 at 9:55pm.

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?

- Math -
**Reiny**, Monday, March 3, 2008 at 10:02pm
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)

- Math -
**tchrwill**, Tuesday, March 4, 2008 at 9:56am
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.

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