Question

Can anyone please solve this question? The answer given is option a ... but how? ?

Answer 1

Let equation of trajectory is,

y=xtana-(gx^2/2u^2cos^2a)

P=Qtana-(gQ^2/2u^2cos^2a)

gQ^2/2u^2cos^2a=Qtana-P

*g/2u^2cos^2a=(Qtana-P)/( Q^2)-(1)

Q=Ptana-(gP^2/2u^2cos^2a)

*g/2u^2cos^2a=(Ptana-Q)/(P^2)-(2)

• from (1) &(2)

(Qtana-P)/(Q^2)=(Ptana-Q)/(P^2)

P^2Qtana-P^3=Q^2Ptana-Q^3

P^2Qtana-Q^2Ptana=P^3-Q^3

PQtana(P-Q)=(P-Q)(P^2+PQ+Q^2)

tana=(P^2+PQ+Q^2)/(PQ)

• therefore,

a=tan^-1[(P^2+PQ+Q^2)/(PQ)]

answer is option (a)