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?

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