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17 Sep 2007 13:39:05 IST
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ramneek kakkar
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17 Sep 2007 15:46:10 IST
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whats d ans
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17 Sep 2007 17:03:38 IST
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Is the angle zero degree with the horizon? In that case only the stone will not come back to the thrower by the influence of the gravity. Considering both vertical and horizontal direction it wont come towards the thrower... Please rate me if i am correct.
18 Sep 2007 18:36:29 IST
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But at other angles excepting 0 degree considering the vertical direction the stone will come towards the thrower Isn't it??
So i think it should be 0 degree with the horizontal.
Beside at 45 degree angle of projection it will have maximum range but in horizontal direction it will come back to the thrower.
If the angle be 0 deg then it will have no vertical motion and hence there being no acc in the horizontal direction the stone will not come back and will proceed having const velocity away frm the thrower!!
So i think it should be 0 degree with the horizontal.
Beside at 45 degree angle of projection it will have maximum range but in horizontal direction it will come back to the thrower.
If the angle be 0 deg then it will have no vertical motion and hence there being no acc in the horizontal direction the stone will not come back and will proceed having const velocity away frm the thrower!!
18 Sep 2007 22:14:48 IST
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the que has two parts
1)that it is always moving away from the thrower
and
2)that we need the the max. angle
ans.
we know that to always keep moving away from the thrower we need to throw it with a minimum velocity of magnitude (11.2km/s+x) so that it reaches infinity
but keeps travelling with a velocity xm/s.Since the escape velocity is irrespective of direction,so the max. possible angle can be 180 degrees or if taken otherwise then 90 degree
taking x in the direction of escape velocity
pls rate
19 Sep 2007 12:09:25 IST
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well u all r wrong
o degree is the minimum angle
if u throw a ball at smaller angles the position vector at every progressive point will be more then the at previous pobut at some big ange
at some point of the motion the position vector will be less then that of the privous point sto we have to find the limiting case at where the position vector will progressively increase
this happens to be the logical ans
but for the exact value help from maths department is needed
at the limiting point the position vector and the velocity vectors will be right angles to each other from the if we take dot product of the velocity vector and the position vector will be zero and we will get the limiting angle
the ans is sin inverse of8/9g
rate if u find these useful
o degree is the minimum angle
if u throw a ball at smaller angles the position vector at every progressive point will be more then the at previous pobut at some big ange
at some point of the motion the position vector will be less then that of the privous point sto we have to find the limiting case at where the position vector will progressively increase
this happens to be the logical ans
but for the exact value help from maths department is needed
at the limiting point the position vector and the velocity vectors will be right angles to each other from the if we take dot product of the velocity vector and the position vector will be zero and we will get the limiting angle
the ans is sin inverse of8/9g
rate if u find these useful
19 Sep 2007 13:17:09 IST
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ya sure
if u connect any point on the projectile motion with the origin it is called the position vector
if a stone is thrown for small angles then the position vector's magnitude progressively increses thats what the question wants
that the stone should always be away from the thrower
for bigger angle it happens that the at some point in motion the position vector's magnitude is less than that of its preceding point
sa the max angle lies in between 45 degrres and 90 degree
here lies the understanding of the problem
now for solving we use the normal projectile equation plus 1 more concept
i.e for the limiting case the velocity vector's direction will be perpendicular to that of the position vector as
1>if it is less than at some point further on the position vector becomes less
2>if its more than no problem but sice we want the limiting case we take to be eqaul to 90 degrre
now for the equation
if we take dot produvt of the position vector and the velocity vector cos90 = o from here and other equations we get the max angle somewhere around 72or somewhere near there i dont know the exzct value
but the method is what i just explianed
if u connect any point on the projectile motion with the origin it is called the position vector
if a stone is thrown for small angles then the position vector's magnitude progressively increses thats what the question wants
that the stone should always be away from the thrower
for bigger angle it happens that the at some point in motion the position vector's magnitude is less than that of its preceding point
sa the max angle lies in between 45 degrres and 90 degree
here lies the understanding of the problem
now for solving we use the normal projectile equation plus 1 more concept
i.e for the limiting case the velocity vector's direction will be perpendicular to that of the position vector as
1>if it is less than at some point further on the position vector becomes less
2>if its more than no problem but sice we want the limiting case we take to be eqaul to 90 degrre
now for the equation
if we take dot produvt of the position vector and the velocity vector cos90 = o from here and other equations we get the max angle somewhere around 72or somewhere near there i dont know the exzct value
but the method is what i just explianed
19 Sep 2007 15:04:57 IST
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But Voldemort projectile path is a parabolic path.. So the distance of any point on the parabola from the point of projection will be more than the preceding point.. Isn't its true. Check the diagram i attached and please tell me where I'm wrong..
20 Sep 2007 11:12:38 IST
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In that case voldemort what ever may be the angle of projection unless it is 0 the path will be parabolic.... So from a certain instant the y-ordinate will start to decrease.... Hence the only method is to keep the y-ordinate fixed and thus projection at 0 deg is necessary. If the projection be 0o+
where 
0 then also in the real case the path will be parabolic and hence the stone will vertically come back towards the thrower... Plz tell me where I am wrong bout my concept|
where 0 then also in the real case the path will be parabolic and hence the stone will vertically come back towards the thrower... Plz tell me where I am wrong bout my concept| 
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