Question

A bomber plane moves horizontally with a speed of 600 m/s and a bomb released from it , strikes the ground in 10 s . What will be the angle with horizontal at which it strikes the ground ?

Answer 1

The bomb is dropped , therefore its horizontal velocity = 600m/s and vertical velocity = 0

In 10 sec,
v = u + at
= u + gt
= 0 + 10x10
= 100 m/s

Horizontal velocity remains unchanged because there is no acceleration in the horizontal.

Therefore,

tan A = ( vertical velocity )/ (horizontal velocity )
= 100/600
= 1/6
Angle with the horizontal, A = tan^-1 (1/6)