Question

A pistol fires a bullet horizontally at 380m/s. The gun is 50.0 meters away from the target. How far below the bullseye will the bullet pierce the target? What is the magnitude and direction of the velocity vector when the bullet strikes the target?

Answer 1

Answer:

8.49 cm below the target;

v = 380.0022 m/s; $\theta=0.2^o$

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, $u_x$ = 380 m/s

Distance between the target and the pistol, $S_x$ = 50 m

Acceleration, $a_x$ = 0 $m/s^2$

Using the following Newton's equation of motion,

$S=ut+\frac{1}{2} at^2$

or, $S_x =u_xt+\frac{1}{2} a_xt^2$

or, $50=(380\times t)+\frac{1}{2}(0\times t^2)$

or, $t=\frac{50}{380} s = 0.1316s$

y-component:

Initial velocity of the bullet, $u_y$ = 0 m/s

Distance the bullet sways from the bullseye = $S_y$

Acceleration, $a_y=g=9.81m/s^2$     (g = acceleration due to gravity)

Again using the same equation used above, we get,

$S_y=u_yt+\frac{1}{2}a_yt^2$

or, $S_y=(0\times t)+\frac{1}{2}gt^2=\frac{1}{2}\times 9.81\times(0.1316)^2=0.0849m=8.49cm$

Now, to find the velocity of the bullet just before it hits the target, we shall use the following equation of motion,

$v^2=u^2+2aS$

or, $v_y^2=0+2\times 9.81\times0.0849 m^2/s^2=1.66m^2/s^2$

or, $v_y=\sqrt{1.66} m/s\approx1.3m/s$

Therefore magnitude of the final velocity of the bullet, $v=\sqrt{v_x^2+v_y^2} =\sqrt{380^2+(1.3)^2} m/s= 380.0022 m/s$

and the direction( in degrees about the x-axis),

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{1.3}{380} )=0.1960^o\approx0.2^o$