6. A rifle is aimed horizontally at shoulder height (1.5 meters above the ground) at a target bullseye 700 meters away. The bullet leaves the barrel with a muzzle velocity of does it hit?1000 m/s.
a. Does it reach the target?
b. If so, where on the target does it strike relative to the center? If not, where on the ground

The motion of the bullet follows a parabolic path. First, we have to determine the time it takes for the bullet to reach the ground. We can do it by considering the vertical motion only, which is a free-fall motion, so we can use the equation

[tex]s=ut+\frac{1}{2}at^2[/tex]

where, taking downward as positive direction,

s = 1.5 m is the vertical displacement of the bullet

u = 0 is its initial vertical velocity

t is the time

[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity

Solving for t,

[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.5)}{9.8}}=0.553 s[/tex]

So, the bullet lands 0.553 s after being shot.

The bullet is fired horizontally at a speed of

[tex]v_x = 1000 m/s[/tex]

So, the horizontal distance covered during this time is

[tex]d=v_x t = (1000)(0.553)=553 m[/tex]

And since the target is 700 m away, the bullet will not reach it.

b) 553 m

As we stated in the previous part, the bullet takes

t = 0.553 s

To land to the ground.

Also, it travels at

[tex]v_x = 553 m/s[/tex]

Therefore, it lands on the ground at a distance of

[tex]d=v_x t=(1000)(0.553)=553 m[/tex]