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Physics 2(Plz Help! :) :) ;))

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A bullet leaving a rifle at an angle of 45 degrees travels a distance of 10 Km determine: A). The muzzle velocity B). What angle would you have to shoot the same bullet to go HALF the distance? Lower angle

  • Physics 2(Plz Help! :) :) ;)) -

    First we need to derive the horizontal distance covered for a muzzle velocity of u, and angle with the horizontal θ.

    The time t the bullet stays in the air is
    Sy=0 = u*sin(θ)*t - (1/2)gt²
    Solving for t gives t=2u*sin(θ)/g.

    Substitute t in the usual formula for horizontal distance:
    Sx=(u*cos(θ))*t
    =2u²sin(θ)cos(θ)/g
    =u²sin(2θ)/g ....(1)

    For the present case,
    Sx = 10,000m
    Solve for in equation (1)
    10000 = u²sin(2*45°)/g
    u=sqrt(10000g/1)=313.05 m/s

    To travel less distance horizontally, we cannot change u, but we can vary θ such that:
    5000 = 313.05&aup2;*sin(2θ)/g
    or
    sin(2θ)=5000g/313.05²=0.5
    asin(0.5)=30° or 120°.

    Therefore θ can be 15° or 60°.

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