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Math Help - Physics Question

  1. #1
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    Physics Question

    1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

    2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

    (Neglect Air Resistance)
    ---

    Velocity in the Y: Sin(40)*2500= 1606 m/s
    Velocity in the X: Cos(40)*2500= 1915 m/s

    Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

    Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

    Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
    t^2= 2(147846)/9.8 = 30173
    t= sqrt(30173)= 174 seconds

    Total time in air= 174+164= 338 seconds

    Horizontal Range: (1915)(338)= 647270
    ---

    Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by Caturdayz View Post
    1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

    2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

    (Neglect Air Resistance)
    ---

    Velocity in the Y: Sin(40)*2500= 1606 m/s
    Velocity in the X: Cos(40)*2500= 1915 m/s

    Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

    Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

    Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
    t^2= 2(147846)/9.8 = 30173
    t= sqrt(30173)= 174 seconds

    Total time in air= 174+164= 338 seconds

    Horizontal Range: (1915)(338)= 647270
    ---

    Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.
    HI

    For (1) ,

    v^2=u^2+2as

    0=(2500\sin 40)^2-2(9.8)(H)

    Greatest height , H = 131752.5 m

    Here , i am not sure what time in the air means .. is it time taken to reach maximum height , or time of flight .

    Anyways , if its time taken to reach max height ,

    v=u+at

    0=(2500\sin 40)-(9.8)t

    t=164 sec

    Then time of flight would be 164 x 2 = 328 since air resistance is neglected .

    Horizontal range , s = vt

    s=(2500\cos 40)(328)

    =628156.4 m
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Caturdayz View Post
    1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

    2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

    (Neglect Air Resistance)
    ---

    Velocity in the Y: Sin(40)*2500= 1606 m/s
    Velocity in the X: Cos(40)*2500= 1915 m/s

    Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

    Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

    Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
    t^2= 2(147846)/9.8 = 30173
    t= sqrt(30173)= 174 seconds

    Total time in air= 174+164= 338 seconds

    Horizontal Range: (1915)(338)= 647270
    ---

    Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.
    For (2) , i assume it still uses the info in (1) .

    s=ut+\frac{1}{2}at^2

    -3=(2500\sin 40)t-\frac{1}{2}(9.8)t^2

    Displacement is negative , then take the non zero value of t .

    Horizontal range , s =(2500\cos 40)(t)

    The t you calculated above .

    For greatest height , you are now just 3 m above where you just fired just now so what would happen to the height of the missle ? neglecting air resistance .
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