A gun can impart a maximum speed of 100m/s to a bullet.A boy holding a stationary target by a horizontal distance of 50 m .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 separation between P and the target is
Supriya Renavikar , 3 Years ago
Grade 11
1 Answer
akshay sinha
Last Activity: 3 Years ago
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=100sinθ1/2cosθ−1/2 g1/2cosθ² → 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°=1.458 m
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