# Maximum range of an object projected from a height h

## 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.

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.

## The Attempt at a Solution

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CAF123
Gold Member

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.

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.

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.

CAF123
Gold Member
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.

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.

CAF123
Gold Member
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.

Yeah, now I understand my mistake. Thanks.