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Help please! simple Physics motion in one dimension problem! Urgent?

  • Thread starter nchin
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  • #1
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Help please! simple Physics motion in one dimension problem! Urgent!?

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 6.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

(a) and (b) i know how to get the answer. but im stuck on c.

Here's what i did:
I chose up as my negative direction (-9.8m/s^2) and down as my positive direction (+9.8m/s^2).

i found the initial velocity to be 15.68m/s
so to find the height of the building i use the formula 'distance = Vo(t) + 1/2at^2'

height of building (distance) = 15.68 (6) + 1/2 (+9.8)(6^2) = 270.48m

my answer is wrong! its suppose to be 82.3m, if i had chose up as my positive direction (+9.8m/s^2) and down as my negative direction (-9.8m/s^2), i would have gotten the right answer. BUT i thought if you stayed consistent with your pos and neg directions you will get the right answer either way but im not getting it.

HELP PLEASE!!!!!!
 

Answers and Replies

  • #2
23
1


i think that what you are doing wrong is that if you want to have down as possitive and up as negative then u have to start out with upwards velocity being negative as well so that
-15.68(6) + 1/2(9.8)(6^2) = 82.32 , and you are correct you can chosse your own frame of reference but you have to make sure that uou are using all the vectors with the right" direction according to you :)
 
  • #3
326
3


A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 6.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

(a) and (b) i know how to get the answer. but im stuck on c.

Here's what i did:
I chose up as my negative direction (-9.8m/s^2) and down as my positive direction (+9.8m/s^2).

i found the initial velocity to be 15.68m/s
so to find the height of the building i use the formula 'distance = Vo(t) + 1/2at^2'

height of building (distance) = 15.68 (6) + 1/2 (+9.8)(6^2) = 270.48m

my answer is wrong! its suppose to be 82.3m, if i had chose up as my positive direction (+9.8m/s^2) and down as my negative direction (-9.8m/s^2), i would have gotten the right answer. BUT i thought if you stayed consistent with your pos and neg directions you will get the right answer either way but im not getting it.

HELP PLEASE!!!!!!
The height of the building is not given here. You have the right initial velocity, which is v_0 ≈ 15.69 m/s [the digit is a bit off, but that is fine with me.]. Now, you will need to substitute that and the value of the gravity [MUST be negative since gravity drags down the projectile] for THIS formula:

s = h_0 + v_0 * t + ½ * a * t²

Make note that h_0 is the initial height. Treat it as the height from the bottom to the top of the building. This is the solution you need to solve for.

This should be fine for you to work out. Remember that at t = 6, s = 0. Find h_0, so you get the answer!
 
  • #4
1,065
10


height of building (distance) = 15.68 (6) + 1/2 (+9.8)(6^2) = 270.48m

my answer is wrong! its suppose to be 82.3m, if i had chose up as my positive direction (+9.8m/s^2) and down as my negative direction (-9.8m/s^2), i would have gotten the right answer. BUT i thought if you stayed consistent with your pos and neg directions you will get the right answer either way but im not getting it
--------
You can get the answer too by taking time equal to (6-1.60-1.60)sec.
 
  • #5
326
3


height of building (distance) = 15.68 (6) + 1/2 (+9.8)(6^2) = 270.48m

my answer is wrong! its suppose to be 82.3m, if i had chose up as my positive direction (+9.8m/s^2) and down as my negative direction (-9.8m/s^2), i would have gotten the right answer. BUT i thought if you stayed consistent with your pos and neg directions you will get the right answer either way but im not getting it
--------
You can get the answer too by taking time equal to (6-1.60-1.60)sec.
That is true. It's important to be consistent with the values you are using for the certain problem. As long as you understand the concept, you are able to solve similar problems.

Here are some good-to-go practice problems:

http://www.physicsclassroom.com/class/vectors/u3l2e.cfm
http://www.physicsclassroom.com/class/vectors/u3l2f.cfm

Good luck!
 

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