A carrom board (4ft x 4ft square ) has the queen at the centre . The queen , hit by the striker moves to the front edge , rebounds and goes in the hole behind the striking line . Find the magnitude of displacement of the queen
(a) from the centre to the front edge,
(b) from the front edge to the hole and
(c) from the centre to the hole
Concept of Physics - 1 , HC VERMA , Chapter - "Physics and Mathematics"
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ANSWER::
In triangle ABC,
tan Ф = x/2 ................. equation 1
In triangle DCE,
tan Ф = (2-x)/4 ................... equation 2
Equating both equation 1 and 2
x/2 = (2-x)/4
2x = 2-x
3x = 2
x = 2/3 ft.
(a) In triangle ABC (by Pythagoras theorem)
AC = √(AB² + BC²) = (2√10)/3 ft.
(b) In triangle CDE ,
DE = 1 - (2/3) = 4/3 ft.
CD = 4ft.
So, by Pythagoras Theorem
CE = √(CD² + DE²) = (4√10)/3 ft.
(c) In triangle AGE , (by Pythagoras Theorem)
AE = √(AG² + GE²) = 2√2 ft.
Hope it helps!
ANSWER::
In triangle ABC,
tan Ф = x/2 ................. equation 1
In triangle DCE,
tan Ф = (2-x)/4 ................... equation 2
Equating both equation 1 and 2
x/2 = (2-x)/4
2x = 2-x
3x = 2
x = 2/3 ft.
(a) In triangle ABC (by Pythagoras theorem)
AC = √(AB² + BC²) = (2√10)/3 ft.
(b) In triangle CDE ,
DE = 1 - (2/3) = 4/3 ft.
CD = 4ft.
So, by Pythagoras Theorem
CE = √(CD² + DE²) = (4√10)/3 ft.
(c) In triangle AGE , (by Pythagoras Theorem)
AE = √(AG² + GE²) = 2√2 ft.
Hope it helps!
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Explanation:
A carrom board (4ft.×4ft.) has the.......here the soln.
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