Solving Rocket Launch at 75 Degrees After 11.2s

[HomerPepsi]

Homework Statement

A rocket is launched at an angle of 75 degrees and accelerates at 21.4 m/s^2. Where is the rocket located (from the point of origin) after 11.2 seconds?

Homework Equations

a=(vf - vi) / t

(vf + vi) / 2 = (xf - xi) / t

where: vf = velocity final, vi = velocity initial, xf = displacement final, xi = displacement initial, and of course a = acceleration and t = time

The Attempt at a Solution

given solution: Xfx = 353.7 m; Xfy = 693.4 m (don’t forget gravity in the y-direction)

X
ax = 21.4cos75 = 5.54
vix = vi cos75
vfx= vi cos75
xfx = ?
xix = 0
t = 11.2

Y
ay = 21.4sin75 = 20.67
viy = vi sin75
viy = ?
xfy = ?
xiy = 0
t = 11.2

any one got any Ideas to the path to the solution, and how do i account for gravity? add/subtract?

thanks

 

[Durden]

I believe your main problem is that you don't have the equation

d = vt + (1/2)at^2 ; where v is initial velocity, t is time, and a is your acceleration.

Once you break the acceleration into it's x and y components, you should be able to get both the x and y distances with the above formula.

Also one thing to note is that the acceleration in the x direction doesn't get effected by gravity, and the acceleration in the y direction is the difference between your original acceleration and gravity. (i.e initial velocity - gravity)