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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

(probably wrong)

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
v=v(initial velocity)+at
v=0+9.8(20s)
v=196 m/s

use another kinematics equation

v^2-V(initial velocity)^2/2a=y

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

Problem 1:

Vo = 0 m/s
a = -9.8 m/s^2
t(up&down)= 20 s

s=Vot+(1/2)at^2
s=(-4,9) * 400 = -1960

height = |s/2| = 980m //

Problem 2:

Vo= 0 (This is obvious)
a= -9,8 m/s^2
t= 3 s

Vf=Vo+at
Vf= (-9,8)*3
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

Sidis

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.

First Problem
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

Second problem:
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
Check:
14.715^2 = (14.715)^2 + 2(-9.81)(0)
Correct.
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
1/2*9.8*100
= 490
thats its

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

490..thats it.
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.