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)required- answer and solution

 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 accⁿ  in the horizontal 

therefore 

tan Θ =  ( vertical velocity )/ (horizontal velocity )  = 1/5

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