An object is thrown with a velocity v at an angle θ with the horizontal. Let the area enclosed by the curve of the projectile with the horizontal be A. For constant value of v the maximum value of A occurs at angle 'a' degrees and it is given by c × g 2 b × v 4 where g is the acceleration due to gravity. Find a + b + c .
Details and assumptions.
1. a,b,c are integers.
2. b contains no factor which is a perfect square.
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A = ∫ 0 x m a x y d x = ∫ 0 2 v s i n ( θ ) / g ( ( v s i n ( θ ) t − g t ² / 2 ) v c o s ( θ ) d t = 3 g 2 2 v 4 sin 3 θ cos θ T h e m a x i m u m o c c u r s w h e n tan θ = 3 , t h a t i s , w h e n θ = 6 0 ° .
actually you don't know how to enclose latex
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A = ∫ 0 x m a x y d x = ∫ 0 2 v s i n ( θ ) / g ( v s i n θ t − g t ² / 2 ) v c o s θ d t
= 3 g 2 2 v 4 sin 3 θ cos θ
The maximum occurs when tan θ = 3 , that is, when θ = 6 0 °
Time to reach highest point is say t. Then t = g V S i n ( a ) H = 2 V S i n ( a ) ∗ t = 2 V S i n ( a ) ∗ g V S i n ( a ) ∴ H = 2 g V 2 S i n 2 ( a ) . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) R a n g e R = 2 ∗ V C o s ( a ) ∗ t = 2 ∗ g V 2 C o s ( a ) ∗ S i n ( a ) ∴ R = 2 ∗ g V 2 C o s ( a ) S i n ( a ) . . . . . ( 2 ) . A = 3 2 ∗ H ∗ R . F r o m ( 1 ) a n d ( 2 ) . . . A = 3 2 ∗ g 2 V 4 C o s ( a ) S i n 3 ( a ) D i f f e r e n t i a t i n g C o s ( a ) S i n 3 ( a ) a n d e q u a t i n g t o z e r o , w e g e t t a n 2 ( a ) = 3 ⟹ a = 6 0 . . . . . A = 3 2 ∗ g 2 V 4 ∗ 2 1 ∗ ( 2 3 ) 3 a = 6 0 , b = 3 , c = 8 . a + b + c = 7 1
Hei,,,I am a new learner,,,,,who you plz say me how A=(2/3)H*R???
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Area of the portion of a parabola bounded by a rectangle dimension HxR = 2/3 x H x R
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If we drop a vertical from the highest point of the projectile take it's intersection with the horizontal origin.
Then equation of curve is of the form y = b − a x 2
Area of the curve(which is above x-axis) with the x-axis is given by :
3 2 b a b
It is easy to note that in our case b = 2 g v 2 s i n 2 θ and a = 2 v 2 c o s 2 θ g (By the equation of trajectory of projectile).
Putting in the values we get A = 3 g 2 2 v 4 s i n 3 θ c o s θ
Calculating d θ d A and equating it to 0 we get θ = 6 0 d e g r e e s
putting θ = 6 0 0 we get :
A = 8 3 g 2 v 4
So a = 6 0 , b = 3 , c = 8 hence a + b + c = 7 1