Solving for Initial Speed and Time in Projectile Motion

[xsc614]

Homework Statement

You shoot an arrow into the air. After 0.9 seconds the arrow has gone straight upward to a height of 30.0 m above its launch point. Ignore air resistance.
(a) What was the arrow's initial speed?
(b) How long did it take for the arrow to first reach a height of 15 m above its launch point?

The Attempt at a Solution

For the first part of the question, I tried multiple things and got wrong answers. By knowing gravity, I changed it from 9.81 m/s2 to 8.829 per .9s, then just added the 30m to 8.829.. which didn't work. I tried assuming constant acceleration, then estimated how far the arrow would have gone in 1 s, and added 9.81 m/s2. I had no idea for the second part, just assumed half the distance would take half the time, .45s was wrong.

Any assistance with this problem would be great.

 

[denverdoc]

need to show eqns, you are aware of. In this instance:
Y=Yo+V*t +1/2 at^2. Assumptions are necessary at times but seems to me you're trying to force fit a solution without proper understanding of the problem. Everyone here has at least tried this approach, so no dishonor in that.

 

[Bitter]

Homework Statement

You shoot an arrow into the air. After 0.9 seconds the arrow has gone straight upward to a height of 30.0 m above its launch point. Ignore air resistance.
(a) What was the arrow's initial speed?
(b) How long did it take for the arrow to first reach a height of 15 m above its launch point?

The Attempt at a Solution

For the first part of the question, I tried multiple things and got wrong answers. By knowing gravity, I changed it from 9.81 m/s2 to 8.829 per .9s, then just added the 30m to 8.829.. which didn't work. I tried assuming constant acceleration, then estimated how far the arrow would have gone in 1 s, and added 9.81 m/s2. I had no idea for the second part, just assumed half the distance would take half the time, .45s was wrong.

Any assistance with this problem would be great.

Let us look at what we know.

We know acceleration is 9.8 m/s because once an object is released then free fall takes place. We also know the height of the object at a certain distance. So, can you find a formula that uses those three variables plus one missing one?

Perhaps x-x(nought)=v(nought)t+1/2at^2

You really should focus more on understanding the problem. It seems to me that you are lacking in understanding the concepts. Go back and review them and try to see if any of it makes more sense. Then go back to this problem and apply them. After you find the answer, restate how you got it so it becomes internal.