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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 May 2007 11:21:24 IST
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a ball falls from the roof of a building .a man standing in front of a one metre high window in the building notes that the ball takes 0.1 seconds to fall from the top to the bottom of the window .the ball continues to fall and strike the floor.on striking the floor the ball rebounds with the same speed with which it hits the floor .if the ball reappears at the bottom of the window 2 seconds after passing the bottom on its way down ,find the height of the building
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16 May 2007 11:31:48 IST
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is the window 1m high from the ground or its length is 1m
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16 May 2007 11:38:37 IST
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Is the answer 5m ??
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16 May 2007 11:48:44 IST
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is answer 6.4 m
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16 May 2007 11:55:00 IST
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ans=20m (approx.) soln: total time to cover the entire height= u/g+0.1+2/2----------u=velocity of ball at the top of window. 1=u/10+1/20 u=9.5 so T=.95+.1+1=2.05 so h =20m (approx.)
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16 May 2007 12:12:08 IST
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nithns ans is perfect, this is approachin d same result wit a small diff...
since it travels 1m in 0.1s, it has a speed of 10m/s at the bottom of the window...
V=10m/s, hence t=1s (to attain this speed)
now frm here it again travels 1s downward(since it has to take the same time to go down to the floor n come bak)
hence the total time taken is 2s..
thus h=1/2at^2
hence.. h= 20m.
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16 May 2007 13:06:36 IST
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Let the velocity at the top of the window be u. Journey through the window : 1 = u(0.1) + (1/2)g(0.1)2 u = 9.5 m/s Hence distance travelled upto the top of the window, x1 = u2/2g = 4.5125 m Distance travelled through the window, x2 = 1 m Velocity at the bottom of the window, v = u + g(0.1) = 10.5 m/s Time taken by the ball to return to the window from the ground = 2 sec Hence time taken to reach the ground from the bottom of window = 1 sec Distance travelled from bottom of window to the ground, x3 =(10.5)1+(1/2)(10)12 x3 = 15.5 Hence height = x1 + x2 + x3 = 4.5125 + 1 + 15.5 = 21.0125 m
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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17 May 2007 12:03:15 IST
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".....the ball reappears at the bottom of the window 2 seconds after passing the bottom on its way down ,find the height of the building ..." See this one.. say a ball with u=20 m/s falls from x m in 1 s time...so v=u + at v= 30 m/s "........the ball rebounds with the same speed with which it hits the floor ......." So, u=30 m/s v=0 so t will be 3 s SO the time is not coming equal when the ball rebounds PLease tell where is my mistake
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18 May 2007 16:27:32 IST
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V WILL NOT BE 0 IT IS VELOCITY IN OPPOSITE DIRECTION HENCE V=-20M/S BECAUSE IN A STRAIGHT LINE MOTION MAGNITUDE OF VELOCITY IS SAME AT A POINT BUT ONLY ITS DIRECTION CHANGES
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18 May 2007 16:55:13 IST
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Let the height of the building be h.
Let the topmost point of the building be A, The top of the window be B, the bottom of the window be C and the bottom most point of the building be D.
Height of the window, L = 1 m
Time taken by the ball to cross the window, T = 0.1 s
Time taken by the ball to travell from C to D and back to D again, T' = 2 s.
Acceleration, a = -g
Let the velocites of the ball at B & C be vA & vB respectively.
Then
vB = vA + aT
 vB = vA - gT ----------------(1)
Also,
vB2 = vA2 + 2a( y)
vB2 = vA2 + 2gL ---------------(2)
Solving equations (1) & (2) for vB &vA
vB = -(gT + 2L/T)/2 = -10.5 m/s (Assuming g = 10m/s2 )
vA = (gT - 2L/T)/2 = -9.5 m/s
Let the distance from A to B be h1, from C to D be h2
vA2 = 2gh1
h1 = 4.5125 m
Now it can be shown that a ball dropped from a given height takes equal times to drop to the ground from that height and to come back to that height from the ground.
Hence time taken to travel from C to D
T' = 1 s
Hence,
y = vB T' + (1/2)aT' 2
-h2 = vB T' - (1/2)gT' 2
h2 = 15.5 m
Hence the height of the building = h1 + L + h2 = 21.0125 m
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19 May 2007 10:42:50 IST
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how did u show that a ball takes equal time to reach the ground from a point andto come back to the point
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19 May 2007 11:01:42 IST
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since it undergoes an elastic collision, we get tat..
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19 May 2007 11:52:36 IST
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let u be the initial velocity with wich the ball crosses the window
now, 1=u*.1 + 1/2*g*(0.1)^2
solving u get u=19/2 m/s
now s1 be the height which the ball covers before reaching the ground.
therefore s1=19/2 * 1 +1/2let u be the initial velocity with wich the ball crosses the window
now, 1=u*.1 + 1/2*g*(0.1)^2
solving u get u=19/2 m/s
now s1 be the height which the ball covers before reaching the ground.
therefore s1=19/2 * 1 +1/2 g 1*1 {time taken by the ball to travell 2s1 distance is 2 sec ,so in 1 sec it travels s1 distance}
solving we get s1=14.5m now let s3 be the distance it travels before reaching the top of the window. v^2=u1^2 + 2gs3 putting v=19/2 and as u1 is 0, we get
s3=4.5 m therefore height of the tower=14.5+1+4.5 m=20 m [i took g=10m/s^2]
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it is not important where u stand, but in which direction u are moving |
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