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OfflinePhychotron
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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.


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Invisibleshroower
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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 x-axis) and you have the equation for the movement relative to the y-axis:
y (relative to the y-axis) = 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 x-axis, but you're interested in the movie along relative to the y-axis, 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 y-axis,
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 x-axis 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 y-axis and x-axis 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 x-axis 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 :smile: maybe I forgot something...


Edited by shroower (10/30/07 10:22 PM)


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Offlinezoombie
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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.


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

*scoffs at basic kinematics problem*

:snub:


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InvisibleDieCommie


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Re: physics anyone? [Re: WhiskeyClone]
    #7578374 - 10/30/07 10:51 PM (13 years, 4 months ago)

Quote:

WhiskeyClone said:
*scoffs at basic kinematics problem*

:snub:


:lol:  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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