# Answer check-derive a general formula for horizontal displacement

## Homework Statement

Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed vi from height, h.

## The Attempt at a Solution

vyf = vyi -gt
t = vyf/g
x = vxi .t = vi cosΘ (t = vyf/g) = [vi^2 sinΘcosΘ]/g

BvU
Homework Helper
The launch is horizontal, so your theta is known. You run into trouble with sin theta, which is 0.

Also, I don't find any dependence on h. Don't you think that if you launch from a high fortress, the projectile covers a bigger distance than if you launch from a small hill ?

Tip: make a drawing!

The launch is horizontal, so your theta is known. You run into trouble with sin theta, which is 0.

Also, I don't find any dependence on h. Don't you think that if you launch from a high fortress, the projectile covers a bigger distance than if you launch from a small hill ?

Tip: make a drawing!

My solution was to find the time taken for the projectile to reach the ground. Then sub that time, t, into the horizontal displacement formula vi cos Θ .t. Is there a reason why this is not viable?

BvU
Homework Helper
The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

I have the drawing already. It's really just the second half of the parabola.

The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

It's not an issue, is it? Since Θ = 0°, then, vi sin(0°) = 0.

BvU
Homework Helper
Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.

Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.

It struck me that the equation y(t)= y(0) + vi.t -0.5gt^2 can be reduced to y(t) - y(0) = h = vi.t - 0.5gt^2. From there I can solve for time. I think this approach is a parallel reasoning to the one in my attempted solution and hence I do not think my answer in the OP is wrong. The only difference would be that using h allow me to express the solution in the intended manner the question demands.

Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.

h = vyi.t - 0.5gt^2
but since yf < yi , then, yf - yi = -h, and vyi = 0 since vi sin 0° = 0
hence, - h = -0.5gt^2
t = SQRT[2h/g] (time taken for projectile to reach the ground)

given time, t = SQRT [ 2h/g], the horizontal displace in this period of time
= vxi.t = vi cos(0°) [SQRT[2h/g]] = vi[SQRT[2h/g]]

BvU