Solving a Rocket's Motion: Max Height & Horizontal Reach

[MissJewels]

Homework Statement

For 6.5 s, a rocket rises in a straight line oriented at 70° from the horizontal axis with a constant acceleration module of 8m/s². Then she is goes into free fall.
Find:
a) the maximum height
b) its horizontal reach

Homework Equations

t = ( v_fv - v_iv)/a
d = v_iht + (0.5)at²
I believe?
y= y_i + v_yot -(0.5)gt²
x= v_xot

The Attempt at a Solution

so I've got these given infos:
t= 6,5
x_i= 0
x_f= ?
y_i= 0
y_f= ?
a= 8 m/s²

a) to find y_f at v_f=0

y= v_o(sin(70))t -(0,5)(9,8)t²

But then, i wonder if i should replace gravity with acceleration, but seeing as were working with y, doesn't that mean i HAVE to use -9,8 as my g (or a) value? From what i understand, I only use 8m/s² when i have to calculate wth the values of x? I'm confused. Someone explain that at least! I have a strong feeling I am heading in the wrong direction with my work so far. Boost me?


b) as for this one, I am guessing i have to find x_f after 6,8 seconds once i find the value of v₀, I'm guessing things will fall into place. But do correct me if I am wrong?

 

[Redbelly98]

The key to this problem is that there are two parts to the motion. The acceleration is different for each part, so you have to do separate calculations for each part:

(1) The acceleration is 8 m/s², in the same direction that the rocket is traveling. So the rocket moves in a 70°-from-horizontal straight line, starting from rest, for 6.5 s. You first need to figure out how far something moves in a straight line with that acceleration, and then figure out the x and y coordinates given that it was traveling at 70°.

(2) The rocket is in free fall, i.e. standard projectile motion with an acceleration of 9.8 m/s² downward. The initial velocity and position are whatever you calculate for final position and velocity at the end of phase (1). This phase lasts until the rocket hits the ground.

 

[MissJewels]

The key to this problem is that there are two parts to the motion. The acceleration is different for each part, so you have to do separate calculations for each part:

(1) The acceleration is 8 m/s², in the same direction that the rocket is traveling. So the rocket moves in a 70°-from-horizontal straight line, starting from rest, for 6.5 s. You first need to figure out how far something moves in a straight line with that acceleration, and then figure out the x and y coordinates given that it was traveling at 70°.

(2) The rocket is in free fall, i.e. standard projectile motion with an acceleration of 9.8 m/s² downward. The initial velocity and position are whatever you calculate for final position and velocity at the end of phase (1). This phase lasts until the rocket hits the ground.

I just took a picture of my work, but here's my attempt at a) the answer is SUPPOSED to be y_f=281 but I am getting 190... what's wrong?

 

[Redbelly98]

I don't agree with some things you wrote there. For example, the rocket is at rest when it is first launched -- that is true of any rocket launch -- so we know v_i is zero. For that reason, your v_i=? statement is puzzling to me. Also I don't think it's right to say that v_f=0 as you did -- there is no reason to think the rocket is at rest after it has been accelerating for 6.5 seconds.

Can you make a sketch showing the two portions of the motion? You'd have a straight line at 70 degrees, and then a curved path that begins at 70 degrees, but curves downward until the rocket hits the ground.

To start the calculations (after you have drawn the sketch!): moving in a straight line and starting from rest, the rocket accelerates at 8 m/s² for 6.5 s. You can use the one-dimensional kinematic equations to answer these two questions: How far does it go? How fast is it going after 6.5 s?

 

[asteriatic]

Dear student,

Your approach is on the right track, but there are a few things that need to be clarified. First, you are correct in using -9.8 m/s^2 as the acceleration due to gravity in the y-direction. This is because we are dealing with a vertical motion, and gravity is acting in the downward direction. The positive value of 8 m/s^2 is used for the horizontal motion because there is no acceleration acting in that direction.

For part a), you are correct in using the equation y = yi + vyi*t + (0.5)*a*t^2. However, your initial velocity in the y-direction (vyi) should be calculated using the initial velocity in the x-direction (vxi) and the given angle of 70 degrees. This can be done using the equation vxi = vi*cos(theta), where theta is the angle of 70 degrees. Once you have calculated the initial velocity in the y-direction, you can use it in the equation above to find the maximum height.

For part b), you are correct in using the equation x = vxi*t. However, make sure to use the initial velocity in the x-direction that you calculated in part a). Also, make sure to use the time of 6.5 seconds, as given in the problem.

I hope this helps clarify your approach. Good luck with your homework!