Deriving projectile motion result

[PsychonautQQ]

Homework Statement

A gun can fire an artillery shell with a speed v_0 in any direction. Show that a shell can strike any target within the surface given by:

(gr)^2 = (v_0)^4 - (2gz)*(v_0)^2

where z is height and r is horizontal distance the target is from the gun.

Homework Equations

Some useful equations that trivially derived:

(v_0*cos(k))^2 = v_0^2 -2g*z_max

z_max = ((v_0*sin(k))^2) / 2g.

The Attempt at a Solution

I'm having problems getting anywhere near the answer. I tried playing around with the first term after substituting (r / (r^2+z^2)^1/2) for cos(k) ( can you even do this? the surface is an arc.)

Anyone want to help guide me here?

 

[ehild]

Homework Statement

A gun can fire an artillery shell with a speed v_0 in any direction. Show that a shell can strike any target within the surface given by:

(gr)^2 = (v_0)^4 - (2gz)*(v_0)^2

where z is height and r is horizontal distance the target is from the gun.

Homework Equations

Some useful equations that trivially derived:

(v_0*cos(k))^2 = v_0^2 -2g*z_max

z_max = ((v_0*sin(k))^2) / 2g.

You need the equation of the trajectory of the shell z=f(r), not only the maximum height and maximum distance. Check "projectile motion" in your lecture notes.

 

[PsychonautQQ]

looks like I have an equation for that.
It can be trivially shown that:

z = tan(k)*x - (g / (2(vcos(k))^2)*x^2.

Do I use tan(k) = z/r
and
cos(k) = r/(r^2+z^2)^1/2
?

 

[PeroK]

Homework Statement

A gun can fire an artillery shell with a speed v_0 in any direction. Show that a shell can strike any target within the surface given by:

(gr)^2 = (v_0)^4 - (2gz)*(v_0)^2

where z is height and r is horizontal distance the target is from the gun.

Homework Equations

Some useful equations that trivially derived:

(v_0*cos(k))^2 = v_0^2 -2g*z_max

z_max = ((v_0*sin(k))^2) / 2g.

The Attempt at a Solution

I'm having problems getting anywhere near the answer. I tried playing around with the first term after substituting (r / (r^2+z^2)^1/2) for cos(k) ( can you even do this? the surface is an arc.)

Anyone want to help guide me here?

I'd start by drawing a diagram of this surface. Note that it reduces to z as a quadratic in r, so should be easy to draw.

Then you might see a simple approach to solving it.

 

[ehild]

looks like I have an equation for that.
It can be trivially shown that:

z = tan(k)*x - (g / (2(vcos(k))^2)*x^2.

Do I use tan(k) = z/r
and
cos(k) = r/(r^2+z^2)^1/2
?

k is the angle the initial velocity v makes with the horizontal. x and z or r and z are the horizontal and vertical coordinates of the target which has to be a point of the trajectory.