Hard
Question
A projectile is thrown with velocity v making an angle θ with the horizontal. It just crosses the top of two poles, each of height h, after 1 second and 3 second respectively. The time of flight of the projectile is
Options
A.1 s
B.3 s
C.4 s
D.7.8 s
Solution
t(OS) = 1 sec
t(OT) = 3
or t(ST) = t(OT) − t(OS) = 3 − 1 = 2 sec
∴ t(SM)
= t(ST) = 1 sec.
∴ t(OM) = t(OS) + t(SM) = 1 + 1 = 2sec.
∴ Time of flight = 2 × 2 = 4 sec. Ans. ″C″
t(OT) = 3
or t(ST) = t(OT) − t(OS) = 3 − 1 = 2 sec
∴ t(SM)
= t(ST) = 1 sec.∴ t(OM) = t(OS) + t(SM) = 1 + 1 = 2sec.
∴ Time of flight = 2 × 2 = 4 sec. Ans. ″C″
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