JEE Main · 2004 · 24 Apr 2004 · Offline · Chapter_pending

t_{1}

and

t_{2}

are the times of flight of two particles having the same initial velocity

u

and range R on the horizontal, then

t_{1}^{2} + t_{2}^{2}

is equal to

Answer: (B)

Solution

Two launch angles give the same range when they are complementary, say

α

and

9 0^{\circ} - α

. Their times of flight are

t_{1} = \frac{2 u s i n α}{g}

and

t_{2} = \frac{2 u s i n ( 9 0 ^{\circ} - α )}{g} = \frac{2 u c o s α}{g}

, so

t_{1}^{2} + t_{2}^{2} = \frac{4 u ^{2}}{g ^{2}} (sin^{2} α + cos^{2} α) = \frac{4 u ^{2}}{g ^{2}}

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