A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. angle at which it strikes the ground will be (g=10m/s)1. tan^{-1} (1/5)2. tan^{-1} (1/2)3. tan^{-1} (1)4. tan^{-1} (5) Ans) the bomb is dropped , therefore its horizontal velocity = 500
and vertical velocity = 0
in 10 sec
v = u + at
= 0 + 10x10
= 100 m/s
horizontal velocity remains unchanged bcoz there''s no acc^{n }in the horizontal
therefore
tan Θ = ( vertical velocity )/ (horizontal velocity ) = 1/5