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Posted by on Sunday, April 29, 2012 at 5:37am.

Two projectiles A and B are projected at the same time in the same vertical plane. A is projected at a height of 2m above the ground making an angle of 30deg with the horizontal. B is projected with the velocity of 20m/s at an angle of 60deg with the horizontal. If they collide. Determine;
i) initial velocity of A (found it=11.5m/s)
ii)horizontal distance moved on the point of collision
iii)the time taken when they collide.

  • physics - , Sunday, April 29, 2012 at 10:47am

    Let’s take as the origin of the coordinate system the point where the projectile B is projected. The point of projectile A start is h = 2 m (above origin).
    For A:
    x(A) =v(oxA)•t =v(oA)•cosα•t,
    y(A) =h +v(oyA)•t - g•t^2/2 =
    =h +v(oA)•sinα•t - g•t^2/2.
    For B:
    x(B) =v(oxB)•t =v(oB) •cosβ•t,
    y(B) = v(oyB)•t - g•t^2/2 =
    =v(oB)•sin β•t - g•t^2/2.
    For collision point:
    x(A)= x(B), and y(A) =y(B), and the same “t”. Then
    v(oA)•cosα•t = v(oB)•cosβ•t, =>
    v(oA)=v(oB)•cosβ/ cosα =20•cos60/cos30 = 11.5 m/s.

    h +v(oA)•sinα•t - g•t^2/2 =
    =v(oB)•sin β•t - g•t^2/2.
    t = h/{v(oB)• sin β - v(oA)•sinα} =
    = 2/{20•sin 60 -11.5•sin30} = 0.17 s.
    y(B) = v(oB)•sin β•t - g•t^2/2 = =20•sin60•0.17 =9/8•(0.17)^2/2 = 2.8 m.
    x(B) =v(oB)•cosβ•t =20•cos60•0.17=1.7 m.
    Coordinates of the collision point are (1.7 m, 2.8m).

  • physics - , Sunday, April 29, 2012 at 11:37am

    thnakss soo much

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