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Projectile Motion Problem-Help Please

  1. Sep 12, 2004 #1
    Projectile Motion Problem-Help Please ASAP!!

    A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 61.9° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 25.5 m away. By how much does the rocket clear the top of the wall?
     
  2. jcsd
  3. Sep 12, 2004 #2

    Tide

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    Technically, it's not a rocket if there is no thrust! :-)

    You might start off by telling what you've done so far.
     
  4. Sep 12, 2004 #3

    HallsofIvy

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    Are we to assume that the "rocket" is not under any force except gravity? Normally a "rocket" has some force of its own and it is a "projectile" that does not.

    Any way, Knowing the initial speed is 75.0 m/s at an angle of 61.9 degrees, you can calculate that the horizontal and vertical components are vx= 75 cos(61.9) and vy= 75 sin(61.9).

    Now that you know the components of the initial velocity do you know formulas for the x and y components of position at time t? At what time will x be 25.5 m? At what that time, what is y?
     
  5. Sep 12, 2004 #4

    Pyrrhus

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    Think in a xy coordinate system...
     
  6. Sep 12, 2004 #5
    i dont have this question, but i have alot of projectile motion questions, and i need all the help i can get.


    can i use x(t) = x(0) + VxT + 1/2at^2
    so.. 25.5 = 0 + VxT + 1/2(a)T^2

    k we know Vx=75 cos(61.9)
    i just didnt calculated that yet.
    we dont know (a) right? or is it -9.8m/s^2 due to gravity?
    we dont know (t), but we can solve for (t) if we know (a)
     
  7. Sep 12, 2004 #6

    Pyrrhus

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    Projectile motion it's just a mix of Free Fall and Constant Velocity Motion. This means for the Y-axis part it'll experience a constant acceleration of g, and for the X-axis part it will experience constant speed.
     
  8. Sep 12, 2004 #7

    ah i kinda get it now...

    can i use x(t) = x(0) + VxT + 1/2at^2
    so.. 25.5 = 0 + VxT + 0
    so solve for t right?


    ok now for the y-part
    y(t) = y(0) + VyT + 1/2at^2
    25.5 = 0 + (75 sin(61.9))T + 1/2(-9.8)t^2
    is there going to be two times for the y-axis part?
     
  9. Sep 12, 2004 #8

    Pyrrhus

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    Well, a particle is moving with constant speed in the x-axis with a time t and at the same time t is moving with constant acceleration in the y-axis.

    In your frist equation you can find the time t when the particle is at x=25.5, just substitute that time t in your y equation to find the Y value the particle will be and minus the 11 meters of the wall and you will have what you're looking for.

    do you see it?
     
  10. Sep 12, 2004 #9

    yea i get it. thanks for the help.
     
  11. Sep 11, 2005 #10
    A rocket is fired at a speed of 75.0m/s from ground level at an angle of 60.0° above the
    horizontal. The rocket is fired toward an 11.0m high wall, which is located 27.0m away. By how much
    does the rocket clear the top of the wall?
    There are several ways to solve this problem. The questions ask about the height of the rocket when it
    reaches the wall 27.0 m away. So we can solve for the time that it takes to travel horizontally to the wall and
    then use that time to find the height of the rocket.
    We need to calculate the x and y components of the initial velocity:
    v0x = v0cos60.0° = (75.0m/s) cos60.0° = 37.5m/s
    v0y = v0sin60.0° = (75.0m/s) sin60.0° = 65.0m/s.
    To find the time of flight, we can use: x = x0 + v0xt + 1/2 ax t2 (assuming to = 0). Once we plug in our values,
    we have 27.0m = 0 + (37.5 m/s)t + 0.
    So t = (27.0 m)/(37.5 m/s) = 0.72 s.
    To find the height at this point, we can use: y = y0 + v0yt + 1/2 at2.
    So y = 0 + (65.0 m/s)(0.72 s) + 1/2 (-9.8 m/s2)(0.72 s)2 = 44.3 m.
    The rocket clears the wall by 44m - 11m = 33m.
     
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