If an arrow on Earth is fired into the air and reaches a hight of 125 m and penetrates the ground where it lands to a depth of 25 cm, what is it's deceleration? Of course the acceleration due to gravity is 9.8 m s, but I'm stumped on how to calculate deceleration?

 

I successfully calculated the final velocity to 49.5 m s^-1.

didn't we have a conversation about this exact question on IRC a while ago?

The final speed is zero. If you mean the speed you calculated is the speed right before it contacts the ground, then you have what will be used as the initital speed in this problem. So, you have the two speeds and the stopping distance was given. What can you do with those three variables? Do you have any equations relating them?

This question is definitely doing the rounds of the physics teachers of world - I have seen it crop up several times recently. It seems a difficult one - because whilst none of the posts I have seen have had trouble working out the speed of contact with the ground many posters haven't made the link to carry on. And there is an implicit assumption that must be made that I think makes the question a little dodgy; unless it is that assumption that the questioner is looking to teach.

what i've seen most is that the people who work out the acceleration of the arrow coming to a stop don't actually believe the result so figure they've done something wrong when in fact they have the correct answer.

didn't we have a conversation about this exact question on IRC a while ago?

 

 

He seems to have asked this question, and received answers, in several places.

 

 

http://www.thescienc...5a61c9207f40b1f

 

Slow learner ?

I successfully calculated the final velocity to 49.5 m s^-1.

 

Personally I find the above line hilarious especially given the fact that he asked this question several times on multiple forums.