Kinematics 1

Classical Mechanics Level 2

The trajectory of a projectile in a vertical plane is y = a x − b x 2 y=ax-b{ x }^{ 2 } , where a a and b b are constants, and x x and y y are respectively the horizontal and vertical distance of the projectile from the point of projection. What is the maximum height attained and the angle of projection from the horizontal?

a 2 16 b ; tan ⁡ − 1 a b \frac { { a }^{ 2 } }{ 16b } ;\quad \tan ^{ -1 }{ ab } a 2 4 b ; tan ⁡ − 1 a \frac { { a }^{ 2 } }{ 4b } \quad ;\quad \tan ^{ -1 }{ a } a 2 8 b ; tan ⁡ − 1 a b \frac { { a }^{ 2 } }{ 8b } \quad ;\quad \tan ^{ -1 }{ \frac { a }{ b } } a 2 b ; tan ⁡ − 1 b \frac { { a }^{ 2 } }{ b } \quad ;\quad \tan ^{ -1 }{ b }

Soumya Shrivastva by Soumya Shrivastva , 20, India

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2 solutions

Swapnil Das
Feb 29, 2016

Comparing the equation y = a x − b x 2 \large y=ax-b{ x }^{ 2 } with the equation of trajectory, y = x tan ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ \large y= x\tan { \theta -\frac { g{ x }^{ 2 } }{ 2{ u }^{ 2 }\cos ^{ 2 }{ \theta } } } , we get

a = tan ⁡ θ \large a = \tan { \theta } or θ = tan ⁡ − 1 a \large \theta =\tan ^{ -1 }{ a } .

Also, b = g 2 u 2 cos ⁡ 2 θ \large b=\frac { g }{ 2{ u }^{ 2 }\cos ^{ 2 }{ \theta } } , which gives

H m a x = u 2 sin ⁡ 2 θ 2 g = sin ⁡ 2 θ 4 b cos ⁡ 2 θ = tan ⁡ 2 θ 4 b = a 2 4 b \large { H }_{ max }^{ }=\frac { { u }^{ 2 }\sin ^{ 2 }{ \theta } }{ 2g } =\frac { \sin ^{ 2 }{ \theta } }{ 4b\cos ^{ 2 }{ \theta } } =\frac { \tan ^{ 2 }{ \theta } }{ 4b } =\frac { { a }^{ 2 } }{ 4b } .

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Since the quadratic equation of trajectory is given , we can approach it as how we normally solve a graph question.

Which means: 1)Find max. height=max. value of y -By using complete a square or differentiate it

2)the angle of the initial velocity to the horizontal - Find it using the result t a n θ = g r a d i e n t o f g r a p h tan\theta=gradient of graph

-Find initial gradient of the graph to find the initial angle when the particle is projected: When x=0, g r a d i e n t = d y d x = a gradient =\frac{dy}{dx}=a

tan ⁡ θ = a \tan\theta=a

Hope it helps=)

Tan LeTian - 5 years, 3 months ago
Ronit Dutta
Jul 9, 2018

At max height maxima would be achieved as the parabola for the trajectory or the path taken by the projectile opens downwards.

Now differentiating both the sides = dy/dx = a - 2bx

At maxima dy/dx=0

therefore, a - 2bx = 0 x = a/2b

substituting value of x in the equation for max height achieved using the given formula. y = a^2/4*b

Now comparing the values of the given equation with the formula for trajectory of a projectile we find tan(theta) = a Thus the initial angle made by the velocity vector is tan^-1 (a).

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