How do I easily solve the following?

A model rocket travels at 88 fps. How long will it take to get to 40 feet high. If we use the equation: h=-16t^2 + vt the answer is 0.5 s and 5 s. This makes no intuitive sense to me as at 0.5 seconds the model rocket should be at 44 feet high.

The problem is unclear. I understand what it's trying to pull out of us, and I can tell you that it's quite poorly worded. Here's what it's trying to tell us:

- - This model rocket is launched by a spring, or a compressed water gun.

The rocket's launch speed off the pad is 88 fps. It has no propulsion after launch, like, say, a model rocket engine. It's just coasting after launch.

- - Where you got the '44-ft' from: That would be correct if the model rocket maintained the same constant speed after launch. But it doesn't.

We know that anything launched like that immediafely begins to slow down, because of gravity. It goes up for a while, until its speed peters out at some peak height. Then it begins to fall, going downward faster and faster until it hits the ground.

- - The height of such an object at any time 't' is H = - 16t² + vt

The ' - 16 ' is 1/2 the acceleration of gravity, and the 'v' is the original speed with which it left the launch pad.

So the equation you have to work with here is H = - 16t² + 88t.

The question asks: When will it be 40 feet high?

So you set H=40, and you solve this quadratic equation for 't':

40 = - 16t² + 88t

or

-16t² + 88t - 40 = 0

I'm sure you can solve the quadratic equation, so we don't have to go all through that part together.

It's a quadratic so it has two solutions: t = 0.5 second and t = 5 seconds.

What does that mean?

Easy: The rocket goes up, passes through 40-ft after 1/2 second, and keeps going up because it hasn't lost all of its speed yet. Eventually, it does run out of speed, hits its peak altitude, stops climbing, starts falling, and passes through 40-ft again on the way down, after 5 seconds into the flight.

Is this scenario easier on your intuitives?