|By Mo222 (Mo222) on Thursday, October 02, 2003 - 08:39 pm: Edit|
A bullet is fired vertically upward and returns to the ground in 20 s. Find the height it reaches.
|By Jason817 (Jason817) on Thursday, October 02, 2003 - 08:47 pm: Edit|
980 m? If its right, I'll explain (unlikely)
|By Euphoria (Euphoria) on Thursday, October 02, 2003 - 08:49 pm: Edit|
im not sure whether this is right but...
first find v using kinematic equation
use another kinematics equation
196^2-0^2/2(9.8 m/s)= 1960m
|By Mo222 (Mo222) on Thursday, October 02, 2003 - 08:51 pm: Edit|
Jason -- I don't know, its an even problem.. =(
Here's another 1 I don't know how to do: ( I know the answer to this 1)
Find the initial and final velocities of a ball thrown vertically upward that returns to the person throwing it 3.0 s later.
|By Mo222 (Mo222) on Thursday, October 02, 2003 - 08:53 pm: Edit|
Euphoria -- I think that it takes 20 s for it to go up AND come back down.
|By Mo222 (Mo222) on Thursday, October 02, 2003 - 08:58 pm: Edit|
Ug Help -- I Cant get any of these!
|By Bard (Bard) on Thursday, October 02, 2003 - 09:05 pm: Edit|
is the final velocity for the second problem 29.4 m/s and initial velocity 0
|By Sidis (Sidis) on Thursday, October 02, 2003 - 09:25 pm: Edit|
Vo = 0 m/s
a = -9.8 m/s^2
t(up&down)= 20 s
s=(-4,9) * 400 = -1960
height = |s/2| = 980m //
Vo= 0 (This is obvious)
a= -9,8 m/s^2
t= 3 s
Vf= -29.4m/s^2 (neg, because it goes down)
Well I guess they are rather easy problems, you just have to substitute the data in the formulas. Hope that helps
|By Jason817 (Jason817) on Thursday, October 02, 2003 - 09:35 pm: Edit|
holy crap I got it right (but I think I did it a different way)...
|By Perry2006 (Perry2006) on Thursday, October 02, 2003 - 09:53 pm: Edit|
My answers are both different from yours guys.
Correct me if I'm wrong.
Displacement is 0 because it falls back to the original position.
X=Vi(t) + .5(-9.81)*t^2
0=vi(20) + .5(-9.81)*(400)
Solve for vi, u get 98.1 m/s
Vf^2 = Vi^2 + 2aX (Note that Velocity is considered 0 when the object reaches its maximum height)
0^2 = 98.1^2 + 2(-9.81)(X)
Solve for X, I have 490.5 = 4.9 * 10^2 m
Again, X (displacement is 0)Thus we have:
X= Vi(t) + .5(-9.81)(t)^2
0= Vi(3) + .5(-9.81(9)
Solve for Vi, you have 14.715 = 15 m/s
Solving for Vf.
Vf = 4.715 + -9.81(3)
Vf = -14.715 = -15 m/s
14.715^2 = (14.715)^2 + 2(-9.81)(0)
As we can see, when the displacement is 0, the Vf and Vi of a free-fall object are the same regardless of the time involved.
|By Sup4 (Sup4) on Friday, October 03, 2003 - 12:59 am: Edit|
or simply s = 1/2at^2 where t = 10
|By Jason817 (Jason817) on Friday, October 03, 2003 - 01:10 am: Edit|
goddamn forgot the 1/2.
|By Zeus (Zeus) on Friday, October 03, 2003 - 02:53 pm: Edit|
as long as gravity is considered to be constant. (the object reaches the ground with the same speed as it was projected. Also time taken to reach the highest point is half the time of flight.)
|By Rashmi (Rashmi) on Saturday, October 04, 2003 - 09:59 am: Edit|
1)Yes you need to take t=10, because the bullet will take 10 secs for going up and 10 for coming down.
2)Initial velocity is 14.7 m/s.
Ok..one question here. I have always been taught that the velocity at the highest point is the final velocity and so in this case it will be 0. Am I right????? or do I have to take the final velocity as the velocity at the point when the ball comes back to the throwers hand.
Report an offensive message on this page E-mail this page to a friend
|Posting is currently disabled in this topic. Contact your discussion moderator for more information.|
|Administrator's Control Panel -- Board Moderators Only|