Projectile Launched at an angle problem.

[KingHenry]

Homework Statement

A spring-loaded cannon aimed at 47 degrees above the horizontal is on the last car of a long train of flat cars. The train has an initial velocity of 54.3 km/h. At the moment the train begins to accelerate forward at 0.325 m/s2 , the cannon fires a projectile at 180 m/s. The cannon points in the direction that the train is moving.
a.) What is the horizontal range observed by a person standing on the ground?
b.) How far on the train from the cannon does the projectile land? Neglect air resistance

Homework Equations

x=Vi*t+ 1/2a(t)^2

The Attempt at a Solution

I know that x component is 180cos 47 and y component is 180sin 47. Not sure where to go from here

 

[MysticDude]

I'm also in this section of my Physics class but I can be of some help.

I found this formula from my book: $$x - x_0 = v_0cos(\theta)t + \frac{1}{2}at^2$$. So I THINK that for the first part, you have to somehow find the t and get the distance.

EDIT: AS I LOOKED AROUND THE SECTION, THERE IS A SPECIAL FORMULA FOR THE HORIZONTAL RANGE

The formula is: $$R = \frac{v_0^2}{g}sin(2\theta_0)$$
Plugging into the formula should give you the horizontal range.
PS. What book are you using, if you are using a book at all?

 

[KingHenry]

The book I am using is holt physics

 

[MysticDude]

Well, you should be able to get the horizontal range, I'm not sure how to do the second part though, I mean I have an idea but I'm not sure if it is correct at all.

 

[tatiana]

you should first scetch yourself a quick visual of what the question is asking and then draw an x l y chart so that you can separate your x and y variables. this makes it much easier to see what you have and what you need to find in terms of x and y
also try using the equaton
$$x - x_0 = v_0cos(\theta)t + \frac{1}{2}at^2$$