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Phychotron
Crazy Mofo
Registered: 06/17/02
Posts: 9,102
Loc: In A Forest Of Colossal F...
Last seen: 9 months, 29 days

physics anyone?
#7578162  10/30/07 09:34 PM (13 years, 4 months ago) 


A rocket is launched at an angle of 57.0° above the horizontal with an initial speed of 98 m/s. It moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s2. At this time its engines fail and the rocket proceeds to move as a free body.
(a) Find the maximum altitude reached by the rocket.
(b) Find its total time of flight.
(c) Find its horizontal range.
 On a mission to prove that the truth gets you no where. They tried the truth, It didn't work. Then they wrote the bible. Only the foolish fear the inevitable.

shroower
Registered: 06/11/06
Posts: 516
Loc: Europe

Re: physics anyone? [Re: Phychotron]
#7578288  10/30/07 10:20 PM (13 years, 4 months ago) 


Thinking the problem without the acceleration of 30 m/s² in the line of motion first (which is a straight line from the start point of the rocket with a 57° angle with xaxis) and you have the equation for the movement relative to the yaxis: y (relative to the yaxis) = 98*sen(57°)*t  0.5*9.8*t²
If you had an acceleration of 30 m/s² along the initial line of motion, well, the initial line of motion is the line passing through the origin of the launch and 57° with the xaxis, but you're interested in the movie along relative to the yaxis, so you decompose the acceleration in the x and y axis, therefore, a_x = 30*cos(57) ~ 16.34 m/s² a_y = 30*sin(57) ~ 25.16 m/s²
Now you can use the acceleration in the yaxis, You have ~25.16 m/s² up, and g ~ 9.8 m/s² down, then you have 15.36 m/s² up for 3 seconds, therefore,
y(t) = 98*sen(57°)*t + 0.5*15.36*t² = 82.19*t + 7.68*t²
The only acceleration in the xaxis direction is the a_x, then, x(t) = 98*cos(57°)*t + 0.5*16.34*t² = 53.37*t + 8.17*t²
OK, you have the equation for the yaxis and xaxis before the engine failure, now you see the movement of the rocket before the engine failure,
Since the rocket moved this way for 3 seconds, you use t = 3 to find the altitude reach by the rocket before the engine failure:
y(3) = 82.19*3 + 7.68*3² = 246.57 + 69.12 = 315.69m
The distance traveled in the xaxis is: x(3) = 53.37*t + 8.17*t² = 233.64m
Now when the engine turns off you have the normal projectile movement for the rocket with just  g ~  9.8m/s² in the direction of y and no acceleration in the direction of x, you can continue from here using y(3) and x(3) as the starting position, with that same angle for this movement and you'll find the maximum altitude.
I hope it helps, but I'm really not sure if it's right maybe I forgot something...
Edited by shroower (10/30/07 10:22 PM)

zoombie
strangler
Registered: 10/19/06
Posts: 242
Last seen: 10 years, 1 month

Re: physics anyone? [Re: Phychotron]
#7578320  10/30/07 10:30 PM (13 years, 4 months ago) 


for highschool projectile motion stuff just break it into 2 parts (going up and coming down) and its pretty straight forward.

WhiskeyClone
Not here
Registered: 06/25/01
Posts: 16,503
Loc: Longitudinal Center of Ca...

Re: physics anyone? [Re: Phychotron]
#7578372  10/30/07 10:50 PM (13 years, 4 months ago) 


*scoffs at basic kinematics problem*
 Welcome evermore to gods and men is the selfhelping man. For him all doors are flung wide: him all tongues greet, all honors crown, all eyes follow with desire. Our love goes out to him and embraces him, because he did not need it. ~ R.W. Emerson, "SelfReliance"

DieCommie
Registered: 12/11/03
Posts: 29,258


Quote:
WhiskeyClone said: *scoffs at basic kinematics problem*
same here. I just got off of work tutoring that shit for 5 hours. Go to physicsforums.com they are pretty good about answering questions.
 


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