Two stones are projected simultaneously with equal speeds from a point on an inclined plane along the line of its greatest slope upwards and downwards respectively. The maximum distance between their points of striking the plane is double that of when they are projected on a horizontal ground with same speed. If one strikes the plane, two seconds later, the other strikes plane, then

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maximum range up the plane $$=R1=u2g(1+$$$$sin$$β$$)---(1)$$ maximum range down the plane $$=R2=u2g(1$$−$$sin$$β$$)---(2)$$ maximum range on ground $$=R=u2g---(3)$$ Given $$R1+R2=2R+R=4R$$ substitute equations (1),(2),(3) $$u2g(1+$$$$sin$$β$$)+u2g(1$$−$$sin$$β$$)=4u2g$$ on solving we get $$cos$$β$$=12$$ ⇒β$$=45---(4)$$  (b) Time of flight $$=T$$ Tup the $$plane=2usin($$α$$1-$$β$$)gcos$$β$$=T1$$ for maximum range α$$1=$$β$$2+$$π$$4$$ substitute in $$T1$$ $$T1=2usin($$β$$2+$$π$$4-$$β$$)gcos$$β$$=2usin($$π$$4-$$β$$2)gcos$$β$$---(5)$$  Tdown the $$plane=2usin($$α$$2+$$β$$)gcos$$β$$=T2$$ for maximum range α$$2=$$π$$4-$$β$$2$$ $$T2=2usin($$π$$4-$$β$$2+$$β$$)gcos$$β$$=2usin($$π$$4+$$β$$2)gcos$$β$$---(6)$$ given $$T2-T1=2$$ substitute equations (5),(6) $$2usin($$π$$4+$$β$$2)gcos$$β$$-2usin($$π$$4-$$β$$2)gcos$$β$$=2$$ ⇒$$u=gcos$$β$$sin$$$$($$π$$4+$$β$$2)-$$$$sin$$$$($$π$$4-$$β$$2)$$ substitute equation (4) $$u=gcos$$$$π$$$$4sin3$$$$π$$$$8-$$$$sin$$$$π$$$$8$$ $$=12.8m/s$$

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