# Help with Projectile motion- equation problem

## Homework Statement

A projectile is launched horizontally with an initial velocity v0 from a height h. If it is assumed that there is no air resistance, which of the following expressions represents the vertical trajectory of the projectile? (A) h–gv0^2/2x^2 (B) h–gv0^2x^2 (C) h-gx^2/2v0^2 (D) h-gx^2/v0^2

d=vi*t+1/2a*t^2

## The Attempt at a Solution

t=d/v; a=g; d=vi*t+1/2g*(x/v)^2-->d=vi*t+(gx^2/2v^2)--->why is this subtracted from h? Also where does d=vi*t go?[/B]

Student100
Gold Member

## Homework Statement

A projectile is launched horizontally with an initial velocity v0 from a height h. If it is assumed that there is no air resistance, which of the following expressions represents the vertical trajectory of the projectile? (A) h–gv0^2/2x^2 (B) h–gv0^2x^2 (C) h-gx^2/2v0^2 (D) h-gx^2/v0^2

d=vi*t+1/2a*t^2

## The Attempt at a Solution

t=d/v; a=g; d=vi*t+1/2g*(x/v)^2-->d=vi*t+(gx^2/2v^2)--->why is this subtracted from h? Also where does d=vi*t go?[/B]

Your equation writing skills are a bit funky, so I might be off base here, for the one question "why is this subtracted from h" draw a picture using vanilla Cartesian coordinates. What direction is the acceleration pointing in? Explain what you're thinking/why you're doing what you're doing as well.

CWatters
Homework Helper
Gold Member
t=d/v; a=g; d=vi*t+1/2g*(x/v)^2-->d=vi*t+(gx^2/2v^2)

You are on the right lines but you should take more care writing equations. You appear to be using d for both the horizontal displacement (in the equation t=d/v) and the vertical displacement (eg in the equation d=vi*t+1/2g*(x/v)2)

--->why is this subtracted from h?

The object starts from the position (0, h). If you put x=0 into your equation does it give the answer y=h ?

Also where does d=vi*t go?

Vi is the initial vertical velocity. What is the initial vertical velocity if the object is "launched horizontally"?

is x the horizontal range of your projectile?

UchihaClan13

haruspex
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2020 Award
is x the horizontal range of your projectile?

UchihaClan13
It would be the horizontal coordinate at an arbitrary point in the trajectory.

It would be the horizontal coordinate at an arbitrary point in the trajectory.
Okay so all we need is an equation of trajectory for the projectile or rather the vertical motion of the projectile
Right???

UchihaClan13

haruspex
Homework Helper
Gold Member
2020 Award
Okay so all we need is an equation of trajectory for the projectile or rather the vertical motion of the projectile
Right???

UchihaClan13
The trajectory is what it asks for. I don't know why it specifies vertical.
Vertical motion could just mean y as a function of t.

Yes nor do i
A projectile has a general trajectory which has both y and x coordinates in it,at a certain instant of time "t"

UchihaClan13

haruspex
Homework Helper
Gold Member
2020 Award
Yes nor do i
A projectile has a general trajectory which has both y and x coordinates in it,at a certain instant of time "t"

UchihaClan13
The term trajectory refers to the shape of the path, without consideration of time.

CWatters
Homework Helper
Gold Member
Perhaps they mention vertical trajectory so you write an equation for y(x) rather x(y) ? Although it's a bit obvious they mean y(x) as they give the answer equation in the problem statement.

i figured the whole thing out
It's easy

did you get an equation connecting y,x,g and v0?

It's easy to see that if x=0 y=h(for the equation of trajectory)
here I consider the point from which the particle is projected as my origin and work accordingly
Now as there's a term for H-something in the options,one can clearly see that the height or the y-coordinate is assumed from the ground to the corresponding x-coordinate
Now all you need is to subtract your equation of trajectory from the initial y-coordinate(which is?/) and voila
:)

UchihaClan13

CWatters