Ahunter 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

B. products  hope it !