Question

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

Answer 1

a) The bullet will not reach the target

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

`s=ut+(1)/(2)at^2`

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

`a=g=9.8 m/s^2` is the acceleration of gravity

Solving for t,

`t=\sqrt{(2s)/(a)}=\sqrt{(2(1.5))/(9.8)}=0.553 s`

So, the bullet lands 0.553 s after being shot.

The bullet is fired horizontally at a speed of

`v_x = 1000 m/s`

So, the horizontal distance covered during this time is

`d=v_x t = (1000)(0.553)=553 m`

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

`v_x = 553 m/s`

Therefore, it lands on the ground at a distance of

`d=v_x t=(1000)(0.553)=553 m`