Which solution do I take?

An object is released into the air at height 17 ft and initial velocity 40 ft/sec. It is caught at a height of 25 feet. Use the vertical motion model to find how long the object is in motion.

I know the equation is 25=-16t^2+40t+17

I solve this and get t=.22 s or t=2.28 s. Which solution do I use? Could there be two answers for how long it is in motion, or do I just take the longer one? I know if I got a negative time, I would reject it, but since I got two positive times, could it be either?

Re: Which solution do I take?

Quote:

Originally Posted by

**twittytwitter** An object is released into the air at height 17 ft and initial velocity 40 ft/sec. It is caught at a height of 25 feet. Use the vertical motion model to find how long the object is in motion.

I know the equation is 25=-16t^2+40t+17

I solve this and get t=.22 s or t=2.28 s. Which solution do I use? Could there be two answers for how long it is in motion, or do I just take the longer one? I know if I got a negative time, I would reject it, but since I got two positive times, could it be either?

The object moves upwards and reaches a height of 25 feet at t=0.22 s. It then proceeds upwards until it reaches its maximum height. Then it starts falling and reaches a height of 25 feet again at t=2.28 s.

So both solutions are right, depending on when the object is caught.