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Maximum range of an object projected from a height h

  • #1

Homework Statement



A particle is projected with an initial speed u from a point at a height h from the horizontal plane. Find the equation for maximum range of the particle.



Homework Equations


R=u^2sin2θ/g

H=(usinθ)^2/2g





The Attempt at a Solution


Let me assume the angle of projection to be θ. Then the total range of the particle can be given as:
Rt=u^2sin2θ/g + √(2h/g)*(ucosθ)

For finding the condition for maximum range I will have to differentiate this expression. But, I'm not able to convert this expression in the form of just one variable.Please help.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
CAF123
Gold Member
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88

Let me assume the angle of projection to be θ. Then the total range of the particle can be given as:
Rt=u^2sin2θ/g + √(2h/g)*(ucosθ)


How did you get this? Should there not be a factor of 2 in the denominator of the first term and I think the second term is more complicated.
 
  • #3
How did you get this? Should there not be a factor of 2 in the denominator of the first term and I think the second term is more complicated.
First term is the range covered in a normal projectile motion and second term is time taken to fall through a height h multiplied by the horizontal velocity.
 
  • #4
3,812
92
First term is the range covered in a normal projectile motion and second term is time taken to fall through a height h multiplied by the horizontal velocity.
Define "normal projectile motion".

As for the question, start by finding out the time taken to reach the ground or the horizontal plane.
 
  • #5
CAF123
Gold Member
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88
First term is the range covered in a normal projectile motion and second term is time taken to fall through a height h multiplied by the horizontal velocity.
Apologies, you are correct in the first term, when I did it I put everything into one expression and overlooked that 2. For the second term though, when the body reaches the same height h above the plane again it has some velocity which is easily found by conservation by energy.
In your attempt, I think you took the velocity to be zero.
 
  • #6
But, only the horizontal component of the velocity would contribute toward range. That's why I took ucosθ instead of u. My aim is to make the expression in terms of just one variable to differentiate the equation and find the maximum value.
 
  • #7
CAF123
Gold Member
2,902
88
But, only the horizontal component of the velocity would contribute toward range. That's why I took ucosθ instead of u.
Yes, but you have that the time taken for the body to fall the distance h is √(2h/g). This assumes the body was just released a height h above the horizontal plane.

My aim is to make the expression in terms of just one variable to differentiate the equation and find the maximum value.
My interpretation is that u and h are known, while θ is the varying parameter. In ths case, you have x= x(θ) and you can find the value of θ that maximises the range.
 
  • #8
Yeah, now I understand my mistake. Thanks.
 

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