[Casualcryptic]

I am having loads of difficulty with this particular word problem. I'll read out the whole thing in case someone feels it's important but I think the real learning gap is probably somewhere in the realm of velocity and distance. I don't think I am making the connection.

A baton twirler tosses a baton into the air. T%he baton leaves the twirler's hand 6 feet above the ground and has an initial velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. The height of the baton can be modeled by h=-16 squared+45t+6 where t is time in seconds and h is height in feet.

There were several parts to the question. The first, graphing I have already done. The second was finding the x an y intercepts and the vertex. I think I got that right but I am not entirely sure. X=2.94,-.12 Y=6 Vertex=(1.41,6) The problem is, there are two other sections to the word problem that just blow my mind.

1) How long does the baton remain in the air?

2) If the twirler misses the catch, how long after the initial toss before the baton hits the ground?

[PolluxCastor]

The formula velocity is v=at
the for distance is d=(1/2)a(t^2).
(check the formula you had listed.)

The easy way is use the velocity and acceleration to calculate the time of max height.
Then use that time to calculate the height above the release point.

Now with that you can calculate the time to the catch point and upon failing to catch, the ground.

See if that helps.