JEE Main · 2004 · 24 Apr 2004 · Offline · Kinematics

A projectile can have the same range

R

for two angles of projection. If

T_{1}

and

T_{2}

be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

(A) $1/ R^{2}$
(B) $1/ R$
(C)R✓ Correct
(D) $R^{2}$

Answer: (C)

Solution

For a fixed launch speed the two complementary projection angles give equal ranges. Their times of flight are

T_{1} = \frac{2 u s i n θ}{g}

and

T_{2} = \frac{2 u c o s θ}{g}

, so their product is

T_{1} T_{2} = \frac{4 u ^{2} s i n θ c o s θ}{g ^{2}} = \frac{2 u ^{2} s i n 2 θ}{g ^{2}}

. The range itself is

R = \frac{u ^{2} s i n 2 θ}{g}

, so the product collapses to

T_{1} T_{2} = \frac{2 R}{g}

, which is directly proportional to the range.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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