Respuesta :
Answer:
8.49 cm below the target;
v = 380.0022 m/s; [tex]\theta=0.2^o[/tex]
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, [tex]u_x[/tex] = 380 m/s
Distance between the target and the pistol, [tex]S_x[/tex] = 50 m
Acceleration, [tex]a_x[/tex] = 0 [tex]m/s^2[/tex]
Using the following Newton's equation of motion,
[tex]S=ut+\frac{1}{2} at^2[/tex]
or, [tex]S_x =u_xt+\frac{1}{2} a_xt^2[/tex]
or, [tex]50=(380\times t)+\frac{1}{2}(0\times t^2)[/tex]
or, [tex]t=\frac{50}{380} s = 0.1316s[/tex]
y-component:
Initial velocity of the bullet, [tex]u_y[/tex] = 0 m/s
Distance the bullet sways from the bullseye = [tex]S_y[/tex]
Acceleration, [tex]a_y=g=9.81m/s^2[/tex] (g = acceleration due to gravity)
Again using the same equation used above, we get,
[tex]S_y=u_yt+\frac{1}{2}a_yt^2[/tex]
or, [tex]S_y=(0\times t)+\frac{1}{2}gt^2=\frac{1}{2}\times 9.81\times(0.1316)^2=0.0849m=8.49cm[/tex]
Now, to find the velocity of the bullet just before it hits the target, we shall use the following equation of motion,
[tex]v^2=u^2+2aS[/tex]
or, [tex]v_y^2=0+2\times 9.81\times0.0849 m^2/s^2=1.66m^2/s^2[/tex]
or, [tex]v_y=\sqrt{1.66} m/s\approx1.3m/s[/tex]
Therefore magnitude of the final velocity of the bullet, [tex]v=\sqrt{v_x^2+v_y^2} =\sqrt{380^2+(1.3)^2} m/s= 380.0022 m/s[/tex]
and the direction( in degrees about the x-axis),
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{1.3}{380} )=0.1960^o\approx0.2^o[/tex]
