The correct option is C a→p, b→r, c→s, d→q
a→p b→r c→s d→q
y=Px−Qx2 (or) (a)
We know path of trajectory of a projectile is,
y=x tan θ−gx22u2 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