What direction would you take answering this question?

[p3nny96]

Homework Statement

Planning a new stunt where he will be shout out of a cannon into the back of a speeding utility vehicle
the sit slopes downhill at 10 degrees and the cannon will launch him at 35Km/Hr at and angle of 30 degrees as shown in the attached picture

Homework Equations

I cannot seem to find a solution I can't think of a method to go about using.

The Attempt at a Solution

a (y) = -g
V (y) = gt + C ---> V (y) = -gt + 35Sin(30) (c given by subbing t=0 and using the given initial velocity in the text given)
r (y) = (-gt^2)/2 +35tSin(30)

I seem to need another piece of information to be able to solve this question?

 

[SteamKing]

Seems like you should first figure out how far the stunt man is going to travel after he's shot from the cannon.

Remember, at some point in his trajectory, the stunt man's vertical velocity will equal zero. How long does this take? How far from the mouth of the cannon is the stunt man then?

There's no magic formula to solve this problem in one step. You have to start with the cannon shot and work through the various steps.

 

[p3nny96]

The problem is that when finding the maximum height and his horizontal displacement is that he doesn't land back on the same ground level it goes below due to the 10 degree slope that the vehicle is traveling down, so finding the time take to max height doesn't really help.

 

[haruspex]

The problem is that when finding the maximum height and his horizontal displacement is that he doesn't land back on the same ground level it goes below due to the 10 degree slope that the vehicle is traveling down, so finding the time take to max height doesn't really help.

You can express the x and y co-ordinates as functions of t, yes? What else connects those co-ordinates? How many unknowns do you have, and how many equations?

 

[SteamKing]

The problem is that when finding the maximum height and his horizontal displacement is that he doesn't land back on the same ground level it goes below due to the 10 degree slope that the vehicle is traveling down, so finding the time take to max height doesn't really help.

Sure it does. As I mentioned, finding the solution is a multi-step process.

Calculating the time to maximum height is just the first step. The trajectory of the stunt man continues after he reaches maximum height, which is figured next. You want to find out how fer downrange the cannon is going to shoot the stuntman, so that you can then calculate how fast the SUV must be traveling to catch up with him.

As Confucius said, "A journey of a thousand miles begins with one step."

 

[haruspex]

Calculating the time to maximum height is just the first step.

I don't see what this is achieving. The step from there to finding where the stuntman lands is no easier than going from launch to land in one step.

 

[SteamKing]

I don't see what this is achieving. The step from there to finding where the stuntman lands is no easier than going from launch to land in one step.

I'm trying to break the problem's solution into manageable chunks for the OP, who was casting about for suggestions.

Obviously, if the OP had no difficulty in formulating his own solution to this problem, then I'm puzzled as to why he sought the advice of strangers in the first place.

 

[haruspex]

I'm trying to break the problem's solution into manageable chunks for the OP, who was casting about for suggestions.

Ok, but it seems to me that finding where it lands after reaching max height is algebraically identical to finding where it lands from the launch data. The extra step of finding max height just seems like additional work.