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Projectile motion problem where the projectile is a rocket

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched at an angle of 53° above the horizontal with an initial speed of 75 m/s. It moves in powered flight along its initial line of motion with an acceleration of 25 m/s^2. At 30 s(seconds) from launch, its engine fails and the rocket continues to move as a projectile.
    (a) What is the rocket's max altitude? (ymax)
    (b) What is the rocket's total time of flight? (tτ)
    (c) What is the rocket's horizontal range? (R/x)

    My given information looks like this: θ= 53°, v1= 75 m/s, a=25 m/s, t= 30s, Δy=0


    2. Relevant equations
    x=vcosθt
    y=vsinθt+.5at^2
    ymax= -(vsinθ)^2/2a

    3. The attempt at a solution
    I tried resolving the velocity into x and y components but i'm not even sure if that's useful information. I also attempted to use the above equations but to no avail... I just need to get this started so that I can finish it. The problem is i'm not sure where to start.
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 23, 2012 #2
    Yes, these formulas are useful to determine the certain components. To find the altitude, you need to use the second formula: y=vsinθt+.5at^2

    Substitute the given values and try to find the max. You can perform the differentiation if you know calculus.

    dy/dt = vsinθ + at

    Then, let dy/dt = 0 and solve for t. This gives you the extrema if a = g = -9.81.
     
  4. Sep 23, 2012 #3
    I found y using the second equation and got 13,046 m... Would that be ymax? I wasn't sure.
     
  5. Sep 23, 2012 #4
    Actually not. You actually solve for t for the dy/dt equation where dy/dt = 0, so we have...

    0 = vsinθ + at
    t = -vsinθ/a

    Now, find the value of t via substitutions. Know that:

    θ = 53°
    v = 75
    a = g = -9.81

    Finally, substitute the t value for y = vsinθt + ½at²
     
  6. Sep 23, 2012 #5
    okay... and when I solve that, would that be the ymax?
    and why are you using -9.81 for the acceleration?
     
  7. Sep 23, 2012 #6
    A rocket is launched at an angle of 53° above the horizontal with an initial speed of 75 m/s. It moves in powered flight along its initial line of motion with an acceleration of 25 m/s^2. At 30 s(seconds) from launch, its engine fails and the rocket continues to move as a projectile.
    (a) What is the rocket's max altitude? (ymax)
    (b) What is the rocket's total time of flight? (tτ)
    (c) What is the rocket's horizontal range? (R/x)

    My given information looks like this: θ= 53°, v1= 75 m/s, a=25 m/s, t= 30s, Δy=0

    ------------------------------------
    Given θ= 53°, v0= 75 m/s, a=25 m/s2, t= 30s

    First part of flight
    v1-v0=at
    v1=825m/s
    ddia=0.5x825x30=12375m

    2nd part of flight.
    Given θ= 53°, v0= 825 m/s, a=-9.8 m/s2, v1=0

    Maximum height is maximum height of 2nd. part plus y component of ddia.
     
  8. Sep 23, 2012 #7
    what is d sub dia?
    like... diagonal distance?
     
  9. Sep 23, 2012 #8
    Okay i think I get it now... thank you!
     
  10. Sep 23, 2012 #9
    Yes
    Sorry, i should have defined it.
    Anyway you got it.
     
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