[blue7]

Hey everyone! I have this physics problem I'm trying to solve, and I'm not sure if I did it right. Could someone tell me if its right?

The problem is:

A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 75 m/s. It moves for 25s along its initial line of motion with an acceleration of 25 m/s^2. At this time, the engines fail, and the rocket proceeds to move as a free body.

a) What is the rocket's maximum altitude?
b) What is rocket's total time of flight?
c) What is the rocket's horizontal range?

My solution:

a) So I know that
a= 25
t=25s (until the engines fail)
v= 75 m/s

1st- I found my vyo= v x sin (x)

So in this case, its
vyo= 75 x sin (53)= 59.89m/s

2nd- Then, I figured since the engine failed, it stopped and started moving downward. Thus, at 25s, it would reach max height. Also, I didn't use g= -10 because its not free body at this point, and gravity isn't factoring into this, so I plugged in a= 25 for g.

So, I used d= vyot + 0.5gt^2 (or in this case, I'm using a for g)

d= (59.89)(25) + (0.5)(25)(625)
d= 1497.25 + 7812.5
d= 9309.75m


b) Now, this is the part I got confused about. I figured that since its now free body, I would use the same distance formula, but use -10 for g now. Then, I thought I would add this value to 25 seconds.

So, -9309.75m = 59.89t + .5 (-10t^2)

0= -5t^2 + 59.89t + 9309.75
0= 5t^2 - 59.89 - 9309.75

(quadratic equation) t= 49.55 s and then 49.55 + 25 is 74.55 seconds.

c) This part is easy, but again I'm not sure if I did a and b correctly.

vx= 75 x cos (53)
= 45.14 m/s

range= vx x t
range = 45.14 x 74.55
range= 3365.187

Thank you sooooo much if you reply to this!

[tickle]

There is no need to post your question again. You are probably not getting any answers because your post involves homework. It is against forum policy to help with homework questions.

Ms tickle

[blue7]

Ooo sorry I'm new and not sure where this would go.

I'll delete the other one if you want.

Im not asking anyone to do my homework either, I just was curious if anyone could see if it was correct or not, because that what this message board seems to be.

[tickle]

No problem, mr. blue7. Glad to have you post for any reason.

[Stratmando]

You gave it thought and were curious about the problem and solution, If you were in school, you would be able to ask a teacher, If using a computer, teachers have to know of resources available online, including forums.
I hope someone will help you, I can't, neither can other poster. Good Luck.
I think you have a valid question.