Maximum speed of the bullet =100m/s

Horizontal distance of the target = 50m

Let the angle of the gun withe horizontal = θ

Horizontal component of velocity =100cosθ

Vertical component of velocity =100sinθ

Equations of motion

Vertical motion y=100sinθt−1/2 gt²

Horizontal motion X=100cosθt

When x=50 m →50=100 cosθt →t =(50/100 cosθ) =1/(2cosθ)

Also when X= 50 m y=0 i.e. grond level

i.e. 0=100sinθt−1/2 gt²

Palcing the value of t gives

0=100sinθ{1/(2cosθ)}−1/2 g{1/(2cosθ)}²

→ 0=50sinθ/cosθ−9.81/{4cos²θ)

→9.81/{4cos²θ)=50sinθ/cosθ

→(sinθ/cosθ)×cos²θ=(9.81/4)/50

→ sinθcosθ=(9.81/200)=0.04905

→ 2×sinθcosθ=2×0.04905

sin2θ=0.09810

2θ=5.629° i.e. θ=2.814°

Let vertical seperation between P and the target =h

The tanθ=h/x =tan 2.814° →h/50 =tan 2.814°

h= 50tan 2.814°=2.458 m