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Q.

Match list I with list II for a projectile.

 List - I List - II
A.For two angles θ and 90θ with same magnitude of velocity of projectione. Pi¯.Pi¯g
B.Equation of parabola of a projectile, y=PxQx2f.Maximum height = 25% of P2Q
C.Radius of curvature of path of projectile projected with a velocity (Pi¯+Qj¯)ms1 at highest pointg.Range = Maximum height
D.Angle of projection θ = Tan-1(4)h.Range is same
 ABCD
1.fhge
2.hfeg
3.egfh
4.eghf

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a

1

b

2

c

3

d

4

answer is B.

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Detailed Solution

For complementary angle of projection, ranges are same.

For Equation of parabola of a projectile, y=PxQx2 :

H=P24Q=25% of P2Q

Radius of curvature of path of projectile projected with a velocity (Pi¯+Qj¯)ms1 at highest point :

R=vx2g=P2g=Pi¯Pi¯g

again, θ=tan1(4HR)=tan1(4)

R=H

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