Q1 Four persons K, L, M, N are initially at the four corners of a square of side d. Each person now moves with a uniform speed v in such a way that K always moves directly towards N and N directly towards K. The four persons will meet at a time
Q2 A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two time of flights is proportional to
1Q2]the 2 angles are x,90-x.
the product of times of flight=[2usinx/g][2ucosx/g] {sin90-x=cosx}
solving, u get t1t2 is proportional to R.
1In question 1, if I got the question correctly, the four people will be at a different position in accordance with velocity vector of the other one. Hence, at some instant, their position will be
Their motion will be something like this, by the time they meet:-
(Sorry for the poor diagram!!!!)
They meet at point O.
Distance (r) travelled by each one of them will be d/√2
time taken will be-
r/v
1In the first questoion their relative velocity along the line joining them is
vapproach= v Cos 90° - v = -v
now their initial separation is = d
time taken fo them to meet is = -d/-v
= d/v
(here minus represents
decrease in separation
1hey injun i have solved their relative velocity alllong the line joining the two particles at any instant.
my solution is correct.
1Yeah, it is, indeed!!
I solved it using t=displacement/velocity(along displacement)
They are one and the same thing, at different instants:)