Area made by projectile

An object is thrown with a velocity v at an angle θ \theta 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 b × v 4 c × g 2 \frac { \sqrt { b } \times { v }^{ 4 } }{ { c\times g }^{ 2 } } where g is the acceleration due to gravity. Find a + b + c . a+b+c.

Details and assumptions.
1. a,b,c are integers.
2. b contains no factor which is a perfect square.


The answer is 71.

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3 solutions

Ronak Agarwal
Jun 4, 2014

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 y=b-a{ x }^{ 2 }

Area of the curve(which is above x-axis) with the x-axis is given by :

2 b 3 b a \frac { 2b }{ 3 } \sqrt { \frac { b }{ a } }

It is easy to note that in our case b = v 2 2 g s i n 2 θ b=\frac { { v }^{ 2 } }{ 2g } { { sin }^{ 2 } }\theta and a = g 2 v 2 c o s 2 θ a=\frac { g }{ 2{ v }^{ 2 }{ cos }^{ 2 }\theta } (By the equation of trajectory of projectile).

Putting in the values we get A = 2 v 4 s i n 3 θ c o s θ 3 g 2 A=\quad \frac { 2{ v }^{ 4 }{ sin }^{ 3 }\theta cos\theta }{ 3{ g }^{ 2 } }

Calculating d A d θ \frac { dA }{ d\theta } and equating it to 0 we get θ = 60 d e g r e e s \theta =60\quad degrees

putting θ = 60 0 \theta ={ 60 }^{ 0 } we get :

A = 3 8 v 4 g 2 A=\quad \frac { \sqrt { 3 } }{ 8 } \frac { { v }^{ 4 } }{ { g }^{ 2 } }

So a = 60 , b = 3 , c = 8 a=60,b=3,c=8 hence a + b + c = 71 \boxed{a+b+c=71}

Similar question Still no solvers to it

Kushal Patankar - 6 years, 4 months ago
Gabriel Lefundes
Jun 15, 2014

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 = 2 v 4 3 g 2 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 = 60 ° . A\quad =\quad \int _{ 0 }^{ x_{ max } }{ ydx } \\ =\quad \int _{ 0 }^{ 2vsin(\theta )/g }{ \left( (vsin(\theta )t\quad -\quad gt²/2 \right) vcos(\theta )dt } \quad \\ =\quad \frac { 2{ v }^{ 4 } }{ { 3g }^{ 2 } } \sin ^{ 3 }{ \theta \cos { \theta } } \\ The\quad maximum\quad occurs\quad when\quad \tan { \theta } =\sqrt { 3 } ,\quad that\quad is,\quad when\quad \theta =60°.

actually you don't know how to enclose latex

Ronak Agarwal - 6 years, 12 months ago

Here's your solution formatted by me :

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 A = \int _{ 0 }^{ { x }_{ max } }{ ydx } \ =\int _{ 0 }^{ 2vsin(\theta )/g }{ \left( vsin\theta t\quad -\quad gt²/2 \right) vcos\theta dt }

= 2 v 4 3 g 2 sin 3 θ cos θ =\frac { 2{ v }^{ 4 } }{ { 3g }^{ 2 } } \sin ^{ 3 }{ \theta \cos { \theta } }

The maximum occurs when tan θ = 3 \tan { \theta } =\sqrt { 3 } , that is, when θ = 60 ° \theta =60°

Ronak Agarwal - 6 years, 10 months ago

Time to reach highest point is say t. Then t = V S i n ( a ) g H = V S i n ( a ) 2 t = V S i n ( a ) 2 V S i n ( a ) g H = V 2 S i n 2 ( a ) 2 g . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) R a n g e R = 2 V C o s ( a ) t = 2 V 2 C o s ( a ) S i n ( a ) g R = 2 V 2 C o s ( a ) S i n ( a ) g . . . . . ( 2 ) . A = 2 3 H R . F r o m ( 1 ) a n d ( 2 ) . . . A = 2 3 V 4 C o s ( a ) S i n 3 ( a ) g 2 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 = 60..... A = 2 3 V 4 1 2 ( 3 2 ) 3 g 2 a = 60 , b = 3 , c = 8. a + b + c = 71 t=\dfrac{VSin(a)}{g} \\ H=\dfrac{VSin(a)}{2} *t=\dfrac{VSin(a)}{2} * \dfrac{VSin(a)}{g}\\\therefore~H=\dfrac{V^2Sin^2(a)}{2g}.......................(1)\\Range~R=2* VCos(a )*t=2*\dfrac{V^2Cos(a)*Sin(a)}{g}\\ \therefore~R=2*\dfrac{V^2Cos(a)Sin(a)}{g}.....(2).\\A=\dfrac{2}{3}*H*R.\\From~ (1)~ and~(2)...A=\dfrac{2}{3}*\dfrac{V^4Cos(a)Sin^3(a)}{g^2}\\Differentiating ~ Cos(a)Sin^3(a)~ and ~ equating~ to~ zero,~~\\we~get~tan^2(a)=3\\\implies ~a=60.....A=\dfrac{2}{3}*\dfrac{V^4*\dfrac{1}{2}*(\dfrac{\sqrt3}{2})^3}{g^2} \\a=60, ~~b=3,~~c=8.~~~a+b+c= \boxed{71}

Hei,,,I am a new learner,,,,,who you plz say me how A=(2/3)H*R???

Sazzad Hossain Rafi - 5 years, 3 months ago

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Area of the portion of a parabola bounded by a rectangle dimension HxR = 2/3 x H x R

Niranjan Khanderia - 5 years, 2 months ago

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