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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??????

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

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