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Question

The path of projectile is represented by y=PxQx2
Column IColumn II(a)Range(p)PQ(b)Maximum height(q)P(c)Time of flight(r)P24Q(d)Tangent of angle of projection(s)2QgP

A
as, bq, cr, dq
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B
ap, br, cq, dq
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C
ap, br, cs, dq
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D
ap, bq, cr, ds
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Solution

The correct option is C ap, br, cs, dq
ap br cs dq
y=PxQx2 (or) (a)
We know path of trajectory of a projectile is,
y=x tan θgx22u2 cos2 θ (b)
Comparing (a) and (b),
p=tan θ (1)
and Q=g2u2cos2θ (2)
Q=g2u2 cos2 θ×sin θsin θ
=g2u2 cos θ. sin θ.tan θ
Q=tan θ(u2 sin 2θg) (3)
Q=(tan θR)
Q=PR

a) Q=tan θR R=PQ

b) H=u2 sin2 θ2g×cos θcos θ×22
=u2(2 sin θ.cos θ)4g×sin θcos θ
=u2(sin2θ)4g×tan θ
H=P24Q

c) T=2u sin θg
T2=4u2 sin2 θg2×22
=4g(u2 sin2 θ2g)×2
=8g×H=8g×P24Q=2P2gQ
T=(2Qg)P

d) tan θ=P

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