R = √(P²+Q²+2PQcosθ)

p² = 2p² (1+cosθ)

1+cosθ = 1/2

Cosθ = -1/2 = cos 120° θ = 120°.

Both A and B will reach the ground simultaneously. Both have zero initial velocity and cover the same height under the acceleration due to gravity.

The horizontal component is constant in a projectile motion. It is not affected by the force of gravity.

At the highest point of projectile, it has only horizontal velocity while the acceleration due to gravity acts vertically downwards. Therefore they are at an angle of 90°.

Time is taken by the bomb to fall through a height of 490m,

t = √(2h/g) = √((2×490)/9.8)=10s

Distance at which the bomb strikes the ground = Horizontal velocity × time = 1km.