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)

 

[thomate1]

As a scientist, my response to this content would be to first clarify some important information. Is the rocket being launched from a stationary position or is it already in motion? Is there any air resistance or other external forces acting on the rocket?

Assuming the rocket is being launched from a stationary position and there is no air resistance, we can use the equations provided to find the location of the rocket after 11.2 seconds.

First, we need to find the initial velocity of the rocket in the x and y directions. We can do this by using the given acceleration and the angle of 75 degrees. The initial velocity in the x direction (vix) is equal to the initial velocity (vi) multiplied by the cosine of 75 degrees. Similarly, the initial velocity in the y direction (viy) is equal to the initial velocity (vi) multiplied by the sine of 75 degrees.

Next, we can use the equation a=(vf - vi)/t to find the final velocity (vf) in the x direction. We know the acceleration (a), the initial velocity (vix), and the time (11.2 seconds), so we can solve for vf. Similarly, we can use the equation (vf + vi)/2 = (xf - xi)/t to find the displacement (xf) in the x direction. We know the final velocity (vf), the initial velocity (vix), and the time (11.2 seconds), so we can solve for xf.

To account for gravity, we need to use the same process in the y direction. We can use the initial velocity in the y direction (viy), the acceleration due to gravity (9.8 m/s^2), and the time (11.2 seconds) to find the final velocity (vf) and displacement (yf) in the y direction.

Once we have both the x and y components, we can use the Pythagorean theorem to find the total displacement (d) of the rocket. This will give us the distance from the point of origin after 11.2 seconds.

In summary, to solve for the rocket's location after 11.2 seconds, we need to find the initial velocities in the x and y directions, use the equations for acceleration, velocity, and displacement to find the final values, and then use the Pythagorean theorem to find the total displacement.