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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jun 2007 17:10:03 IST
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A boy throws a tennis ball with a velocity of 20 m/s at an angle  with the horizontal. The wind imparts a force of 0.8N horizontally such that the ball returns to the starting point. For this to happen, tan should be equal to ______ .
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Will nip in at times to solve problems :)
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see, the projectile is in the air till it comes down. to find the time of flight: so equating vertical displacement S to 0, 0 = vsin  t - (1/2)*g*t 2so, t = 2vsin / g now, for this time the wind exerts force. given that horizontal displacement is also 0, 0 = vcos t - (1/2)*0.8*t 2so, t = 2vcos / 0.8 equatingboth the times, 2vsin /g = 2vcos /0.8 so, tan = g/0.8 tan = 5g/4 RATE if useful...
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<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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28 Jun 2007 19:01:13 IST
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The mass of the ball is required in the problem.
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28 Jun 2007 19:01:53 IST
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@ Lastmin
The value of tan is dimensionally incorrect. You have taken the horizontal acceleration to be 0.8 whereas it is the force.
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28 Jun 2007 19:11:10 IST
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see in hcv
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28 Jun 2007 19:18:20 IST
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Sorry... Mass of the ball is 1 kg.
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28 Jun 2007 19:27:51 IST
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All right. Then here it goes
The wind is clearly in opposition to the horizontal velocity of the ball. The acceleration due to wind is 0.8 in a direction opposite to the X component of the initial velocity of the ball.
As the vertical displacement is zero,
20sint - (1/2)gt2 = 0
t = 40sin /g
As the horizontal displacement is zero,
20cost - (1/2)at2 = 0
40cos = (0.8)40sin/g
tan = 5g/4
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28 Jun 2007 19:33:02 IST
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i have used 0.8 N force by the wind as the acceleration assuming the weight of the ball was 1kg... @elessar_iitkgp 5g/4 is only in magnitude (coz ive assumed that wt of ball is 1 kg... check my soln carefully, ive used acc=0.8)
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Always be just and unbiased in your opinions and also say them aloud, even if it means the majority of the world to go against you!!
There will be still some left, who will approve of your character and one of them is God!!!
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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28 Jun 2007 19:50:03 IST
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@Lastmin
Yes you have taken acceleration as 0.8. Sorry for the remark
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28 Jun 2007 19:54:05 IST
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Initial velocity = -vcos$
Thus in time period 2vsin$/g
a=0.8 m/s sq horizontal
s=0 --> velocity =0
u=at
v cos$ =0.8.2vsin$/2g
tan $ =5g/4 Ans
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28 Jun 2007 20:55:47 IST
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actually, all of u seem to be wrong.
I just found the original question paper, and the data are as follows:
mass of the ball = 10g. (This part of the question is soiled, so i am not sure if its kg or g.)
inital velocity is the same, ie 20 m/s.
options are tanQ = 0.2, 0.4, 0.8, 1.0...
I tried this sum with your approach, taking the mass to be both in gms and in kgms, but got something else as the answer.
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28 Jun 2007 21:55:46 IST
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Experts?
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29 Jun 2007 19:07:32 IST
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Why doesn't any expert answer this????
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29 Jun 2007 20:03:20 IST
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Experts!!!
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30 Jun 2007 20:48:34 IST
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