Answer:
8.49 cm below the target;
v = 380.0022 m/s; ![\theta=0.2^o](https://tex.z-dn.net/?f=%5Ctheta%3D0.2%5Eo)
Explanation:
Let us divide this problem into it's x- and y-components (horizontal = x and vertical = y)
x-component:
Initial velocity of the bullet,
= 380 m/s
Distance between the target and the pistol,
= 50 m
Acceleration,
= 0 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Using the following Newton's equation of motion,
![S=ut+\frac{1}{2} at^2](https://tex.z-dn.net/?f=S%3Dut%2B%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
or, ![S_x =u_xt+\frac{1}{2} a_xt^2](https://tex.z-dn.net/?f=S_x%20%3Du_xt%2B%5Cfrac%7B1%7D%7B2%7D%20a_xt%5E2)
or, ![50=(380\times t)+\frac{1}{2}(0\times t^2)](https://tex.z-dn.net/?f=50%3D%28380%5Ctimes%20t%29%2B%5Cfrac%7B1%7D%7B2%7D%280%5Ctimes%20t%5E2%29)
or, ![t=\frac{50}{380} s = 0.1316s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B50%7D%7B380%7D%20s%20%3D%200.1316s)
y-component:
Initial velocity of the bullet,
= 0 m/s
Distance the bullet sways from the bullseye = ![S_y](https://tex.z-dn.net/?f=S_y)
Acceleration,
(g = acceleration due to gravity)
Again using the same equation used above, we get,
![S_y=u_yt+\frac{1}{2}a_yt^2](https://tex.z-dn.net/?f=S_y%3Du_yt%2B%5Cfrac%7B1%7D%7B2%7Da_yt%5E2)
or, ![S_y=(0\times t)+\frac{1}{2}gt^2=\frac{1}{2}\times 9.81\times(0.1316)^2=0.0849m=8.49cm](https://tex.z-dn.net/?f=S_y%3D%280%5Ctimes%20t%29%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.81%5Ctimes%280.1316%29%5E2%3D0.0849m%3D8.49cm)
Now, to find the velocity of the bullet just before it hits the target, we shall use the following equation of motion,
![v^2=u^2+2aS](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2aS)
or, ![v_y^2=0+2\times 9.81\times0.0849 m^2/s^2=1.66m^2/s^2](https://tex.z-dn.net/?f=v_y%5E2%3D0%2B2%5Ctimes%209.81%5Ctimes0.0849%20m%5E2%2Fs%5E2%3D1.66m%5E2%2Fs%5E2)
or, ![v_y=\sqrt{1.66} m/s\approx1.3m/s](https://tex.z-dn.net/?f=v_y%3D%5Csqrt%7B1.66%7D%20m%2Fs%5Capprox1.3m%2Fs)
Therefore magnitude of the final velocity of the bullet, ![v=\sqrt{v_x^2+v_y^2} =\sqrt{380^2+(1.3)^2} m/s= 380.0022 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%20%3D%5Csqrt%7B380%5E2%2B%281.3%29%5E2%7D%20m%2Fs%3D%20380.0022%20m%2Fs)
and the direction( in degrees about the x-axis),
![\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{1.3}{380} )=0.1960^o\approx0.2^o](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7Bv_y%7D%7Bv_x%7D%29%3Dtan%5E%7B-1%7D%28%5Cfrac%7B1.3%7D%7B380%7D%20%20%29%3D0.1960%5Eo%5Capprox0.2%5Eo)