Finding the angle of projection and its speed in 2D Kinetics

In summary, the equation for the trajectory of a projectile in a vertical plane is y = √3 x - (1/5)x2, where x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The angle of projection can be found from the gradient of the graph at what point. The speed of projection can be found by using the equation u2Sin2(theta)/g.
  • #1
dahano
10
0

Homework Statement

: The trajectory of a projectile in a vertical plane is y = √3 x - (1/5)x2, where x and y are respectively horizontal and vertical distances of the projectile from the point of projection. Find the angle of projection and speed of projection.[/B]

Homework Equations

: 2-Dimensional Equations and maybe calculus.[/B]

The Attempt at a Solution

: Honestly, I can't even understand that how do I begin with this question. I've tried some wild attempts but they landed me no where :-( Can you please guide me with the solution to this question? [/B]
 
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  • #2
dahano said:
I've tried some wild attempts but they landed me no where :-

Please post these attempts .

However I still give a hint :

What does dy/dx represent ?
What is the formula for Range ?
 
  • #3
The angle of projection can be found from the gradient of the graph at what point?
 
  • #4
I figured out the formula for Ran
Qwertywerty said:
Please post these attempts .

However I still give a hint :

What does dy/dx represent ?
What is the formula for Range ?

I figured out the formula of range as: u2Sin2(theta)/g

Can you tell me that how would it be done further?
 
  • #5
andrevdh said:
The angle of projection can be found from the gradient of the graph at what point?
Sorry to ask a stupid question but what do you mean by point? Can you please show me how would it be solved?
 
  • #6
dahano said:
I figured out the formula of range as: u2Sin2(theta)/g

Can you tell me that how would it be done further?

Here - dy/dx at x = 0 represents tan(θ) ( where θ is angle of projection ) .

Find range , and then using formula for range find u .
 
Last edited:
  • #7
What does the trajectory or path of the projectile look like?
What is the equation which describes the path or trajectory?
 
  • #8
dahano said:
I figured out the formula of range as: u2Sin2(theta)/g
Please attach your working for this.
 

Related to Finding the angle of projection and its speed in 2D Kinetics

What is the angle of projection in 2D Kinetics?

The angle of projection in 2D Kinetics refers to the angle at which an object is launched or thrown. It is measured with respect to the horizontal plane.

How is the angle of projection calculated?

The angle of projection can be calculated using trigonometric functions, such as sine, cosine, and tangent. It can also be determined by using vector analysis.

What factors affect the angle of projection?

The angle of projection is affected by factors such as the initial velocity of the object, the height at which it is launched, and the angle at which it is released.

What is the speed of projection in 2D Kinetics?

The speed of projection in 2D Kinetics is the initial velocity at which an object is launched or thrown. It is a vector quantity and is measured in meters per second (m/s).

How is the speed of projection calculated?

The speed of projection can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. It can also be determined using vector analysis.

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