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A rocket rises straight at a 70 degree angle then falls.using its acceleration,find

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    t = ( vfv - viv)/a
    d = viht + (0.5)at2
    I believe?
    y= yi + vyot -(0.5)gt2
    x= vxot

    3. The attempt at a solution
    so ive got these given infos:
    t= 6,5
    xi= 0
    xf= ?
    yi= 0
    yf= ?
    a= 8 m/s2

    a) to find yf at vf=0

    y= vo(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, doesnt 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 im heading in the wrong direction with my work so far. Boost me?


    b) as for this one, im guessing i have to find xf after 6,8 seconds once i find the value of v0, I'm guessing things will fall into place. But do correct me if im wrong?
     
  2. jcsd
  3. Nov 8, 2011 #2

    Redbelly98

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    Re: A rocket rises straight at a 70 degree angle then falls.using its acceleration,fi

    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 travelling 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.
     
  4. Nov 9, 2011 #3
    Re: A rocket rises straight at a 70 degree angle then falls.using its acceleration,fi

    I just took a picture of my work, but heres my attempt at a) the answer is SUPPOSED to be yf=281 but im getting 190... whats wrong?

    EB988483.jpg
     
  5. Nov 9, 2011 #4

    Redbelly98

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    Re: A rocket rises straight at a 70 degree angle then falls.using its acceleration,fi

    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 vi is zero. For that reason, your vi=? statement is puzzling to me. Also I don't think it's right to say that vf=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?
     
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