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13 August, 04:19

A hunter aims directly at a target (on the same level) 75.0 m away.

(a) If the bullet leaves the gun at a speed of 180 m/s, by how much will it miss the target?

(b) at what angle should the gun be aimed so as to hit the target?

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  1. 13 August, 07:11
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    From above data we have given

    initail velocity=180m/s

    distance of target = 75.0

    Time=t=75/180sec=0.416

    bullet acceleration=a=9.8m/s^2

    u=0

    s=distance covered=?

    now putting in the formula

    s=ut+1/2at^2

    s=0*0.416+0.5 (9.8) (0.416) ^2

    s=0+0.85

    so bullet will miss the target 0.85m

    Now finding the angle

    T = (2 u Sin θ) / g

    and horizontal range is equal to=R

    where R = (u² Sin 2θ) / g

    u is the initial velocity, θ the angle of elevation, g is the acceleration due to gravity

    75 = (180^2 Sin 2θ) / 9.8

    75 = 32400 * Sin 2θ / 9.8

    Sin 2θ = (75 * 9.8) / 32400 = 0.0227

    2θ = 1.3°

    θ = 0.65 °

    so angle gun should be aimed to hit the target is 0.65
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