2D Kinematics Problem involving subtended angle

[RjD12]

Homework Statement

Mortar crew is near the top of a steep hill. They have a mortar. They angle this mortar at an angle of \theta = 65° . The crew fires a shell at a muzzle velocity of 228 ft/sec (69.5 m/s). How far down the hill does the shell strike if the hill subtends an angle of \phi = 32° from the horizontal?

How long will the mortar remain in the air?
How fast will the shell be traveling when it hits the ground?

Relevant diagram: http://imgur.com/apimguh

Homework Equations

Kinematic Equations:

X= X_o + V_oxt

Y= Y_o + V_oyt - (1/2)at²

The Attempt at a Solution

First off, I'm not expecting to get all of my questions answered. I just need a little push.

I'm not sure where to start off at here. The fact that there are two angles here confuses me in regards to how they work in the equations.

I can say that V_ox = 69.5cos(65) and that V_oy = 69.5sin(65).

I'm really thrown off by the way the angles work here, and whether the distance works with a simple range equation. Any tips on where to start?


edit: Additionally, I was given the equation d = V_o + (1/2)at² as a hint for this. Isn't this wrong though, seeing as how the velocity should be multiplied with time?

 

[Simon Bridge]

Always start with a sketch.
I'll start you off:

A "mortar" in this example refers to the mortar tube or gun depending on the period.
Think of it as a kind of cannon. Draw a dot on your page to represent the position of the mortar.

The mortar is sitting on the side of a hill.
The hill has a slope - given as below the horizontal.
So draw in a (dotted) horizontal line through the mortar and figure out which way is "below" - measure the angle in that direction and draw in a bold line through the mortar indicating the slope of the hill.

The angle of the mortar is above the horizontal - you know how to do that right?
So draw an arrow for the initial velocity v at the correct angle.

The mortar shell will go in a parabola until it hits the side of the hill - which keeps going down at the same angle.

You should be able to take it from there.

 

[BvU]

Strange that the picture seems to have been posted before Simon commented. It looks excelent.
I checked your earlier thread and I think you only need a small nudge.

The hint you got is suspicious, because d = Vo + (1/2)at2 is dimensionally incorrect -- as you note. Kinematic relations are OK. Look at them in a math fashion: you still have three unknowns (x, y, t) so something else is needed (for example, in flatland: y(t)=0 when poof).

In this exercise it is the slope of the hill, that provides you with a relationship between X(t) and Y(t) at boom time. If you express it deftly in terms of d and $\phi$ you end up with only two unknowns (d and t_strike) in two equations. Bob's your uncle...

 

[Simon Bridge]

Hmmm... looks like I didn't see it.
Either that or it was edited in while I was doing something else.
I don't think I took a half-hour to type the reply but I sometimes have several threads up in different tabs.

Looking at the diagram:
Basically the projectile ends up at a different height than it started at - this height depends on the horizontal distance traveled.

You could think of it as the intersection of a parabola with a line - do you know how to get the gradient of a line from the slope angle?

Once you have the (x,y) coordinates of the endpoints you can get d.

I had a more elaborate hint but I'll reinforce BvU at this point :)

 

[azizlwl]

It is 2D Kinematics. The time for the shell to reach horizontally at D(x) is equal to time it reaches D(y).