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

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 -
**Elena**, Sunday, April 29, 2012 at 10:47amLet’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 -
**ridhi**, Sunday, April 29, 2012 at 11:37amthnakss soo much

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