DISTANCE travelled by the projectile

[judas_priest]

Homework Statement

What is the DISTANCE traveled by the projectile?
I know how to calculate the displacement, but this includes both x and y components, so how do I calculate the distance? I'm sure it must include both height and vertical displacement components

Homework Equations

The Attempt at a Solution

I have no clue really. Everything I try to do, it just gives me the displacement. I just don't know how to get the 'h' factor in? Do I use pythagaros? Because it doesn't apply for a curved surface.

 

[STEMucator]

Homework Statement

What is the DISTANCE traveled by the projectile?
I know how to calculate the displacement, but this includes both x and y components, so how do I calculate the distance? I'm sure it must include both height and vertical displacement components

Homework Equations

The Attempt at a Solution

I have no clue really. Everything I try to do, it just gives me the displacement. I just don't know how to get the 'h' factor in? Do I use pythagaros? Because it doesn't apply for a curved surface.

Wait are you trying to find the distance the object flew in total or are you trying to find its horizontal displacement ( I'm sure it's displacement though ).

With that in mind, to get the horizontal displacement of an object use :

$$\vec{Δd}_H = \vec{v}_H Δt = \frac{ \vec{v}_R^2 sin(2θ) }{ \vec{a} }$$

depending on which information you have been given or have been able to retrieve.

 

[CAF123]

With that in mind, to get the horizontal displacement of an object use :

$$\vec{Δd}_H = \vec{v}_H Δt = \frac{ \vec{v}_R^2 sin(2θ) }{ \vec{a} }$$

What is ##\vec{v}_R?## That equation does not make sense. You cannot square a vector or divide by one.
@judas_priest: Distinguish between horizontal and vertical motion. Write the equation for the displacement of the body in each direction. Then find the horizontal displacement in terms of known quantities by eliminating variables between the two equations.

 

[STEMucator]

What is ##\vec{v}_R?## That equation does not make sense. You cannot square a vector.
@judas_priest: Distinguish between horizontal and vertical motion. Write the equation for the displacement of the body in each direction. Then find the horizontal displacement in terms of known quantities by eliminating variables between the two equations.

##\vec{v}_R## is the resultant velocity ( Hypotenuse ).

EDIT : Yeah you're right, It should be :

$$\frac{v_R^2 sin(2 \theta)}{a}$$

without the vector quantities.

 

[Curious3141]

Most elementary projectile mechanics question limit themselves to asking about the maximum horizontal distance traveled by the projectile, also known as the range. This is very easy to work out, and the well-known formula for a projectile launched on a flat surface has already been quoted. For a projectile launched from a height, the answer is quite simple to work out from first principles.

But if the question is actually asking for the total distance traveled by the projectile on its curved path, it's not impossible. Tedious, but not impossible. The distance is simply ##\displaystyle \int_0^T v(t)dt##, where ##T## is the total time of flight, and ##v(t)## is the instantaneous speed of the projectile expressed as a function of time.

Using the resultant of the horizontal and vertical components of velocity at time ##t##, it is possible to get an expression for ##v(t)##. Integrating that is a little tough, and a hyperbolic trig substitution is probably the quickest way. And simplifying the resulting mess is daunting. But an answer can be derived fairly quickly.