Hard
Question
A projectile is projected at an angle a (> 45o) with an initial velocity u. The time t at which its horizontal component will equal the vertical component in magnitude:
Options
A.t = u/g ( cos α − sin α)
B.t = u/g ( cos α + sin α)
C.t = u/g (sin α − cos α)
D.t = u/g (sin2 α − cos2 α)
Solution
MCQ
Vx = Vy Now, we could have choose n coordinate axis as
⇒ Time at which Vx = Vy is what we are solving
Now Vx = u cos α
Vy = u sin α − gt
⇒ u cos α = u sin α − gt {∵ Vy = Vx} ; at t = t1 (say)
⇒ t1 = (sin α − cos α) ″C″ Ans
Also when Vy ≡ Vx {i.e., when we choose ′y′ axis as − y′} at t = t2 (say)
− u cos α = u sin α − gt2
⇒ t2 = u/g (sin α + cos α) ″B″ Ans
Vx = Vy Now, we could have choose n coordinate axis as
⇒ Time at which Vx = Vy is what we are solving
Now Vx = u cos α
Vy = u sin α − gt
⇒ u cos α = u sin α − gt {∵ Vy = Vx} ; at t = t1 (say)
⇒ t1 = (sin α − cos α) ″C″ Ans
Also when Vy ≡ Vx {i.e., when we choose ′y′ axis as − y′} at t = t2 (say)
− u cos α = u sin α − gt2
⇒ t2 = u/g (sin α + cos α) ″B″ Ans
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