JEE Main · 2005 · 08 May 2005 · Offline · Kinematics

A projectile can have the same range

R

for two angles of projection. If

t_{1}

and

t_{2}

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

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

Answer: (D)

Solution

Two complementary angles

θ

and

9 0^{\circ} - θ

give the same range, so 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}

. Multiplying them,

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}}

. Since the range is

R = \frac{u ^{2} s i n 2 θ}{g}

, this is simply

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

, so the product of the flight times is directly proportional to

R

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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