Any Physics people here? Help solve this.

[Imaginer]

Problem!

 

[Imaginer]

I have a bit of a problem starting off.

 

[Viper GTS]

Been a while since I've done anything like this, but this is my initial thought:

Relate the position of the tank to time, & the horizontal position of the projectile to time. Find out where the two are equal, & then plug that time (or distance) back into the equation for the vertical position of the projectile to determine theta.

Viper GTS

 

[Imaginer]

But it needs g.

Here is what I have so far.

with time being equal for both tank and shell.

((2/a)(x-xo-vxo))^(1/2) = (x-xo)/vxo

 

[Viper GTS]

Your equation for the vertical position of the shell will have to inclue g to take into account gravitational pull on it. That's where you'll get your g.

Viper GTS

 

[DataFly]

Here's how I did it:

Distance = d
Gravity = g
Tank's Acceleration = a
You know that the vertical speed (Vv) is equal to the velocity (V) times the sin of theta (@). Vv = V * sin @
The horizontal speed of the shell (Vh) is equal to V * cos @. Vh = V * cos @
At the midpoint in time Vv is 0 (it is at the apex), so you can enter 0 for Vfinal in the equation Vfinal = Vinitial + acceleration * time. 0 = (V*sin @) + g*t
When you solve for t you get t = (-V*sin @)/g.
Double this to get the total time (time of ascent + time of descent). t = (-2V*sin @)/g

The horizontal distance can be found by entering the values for Vh and t. d = (V*cos @)*((-2V*sin @)/g)
This is equal to the distance the tank travels, which can be found using the equation d =(1/2)a*(t^2)

Since the distances are equal, you can create the following equation:
(V*cos @)*((-2V*sin @)/g) = (1/2)a*(t^2)

When I solved for @ I got @ = tan^(-1) (-g/a)

I apologize if this is wrong, but I really don't think it is.
Now I need to get to bed so I can take my physics test tomorrow.