Question

A gun can impart a max speed of 100m/s to a bullet.a boy holding this gun is seperated from a stationary target by a horizontal distance of50m.in order to hit the target in minimum time, the barrel of the gun is directed towards a point P,vertically above the target.the vertical seperation between P and the target is??????

Answer 1

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

Answer 2

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