Question

Consider the motion of a bullet that is fired from a rifle 6 m above the ground in a northeast direction. The initial velocity of the bullet is (200, 200, 0) . Assume the​ x-axis points​ east, the​ y-axis points​ north, the positive​ z-axis is vertical​ (opposite g), the ground is​ horizontal, and only the gravitational force acts on the object.
a. Find the velocity and position vectors for t >_0.

Answer 1

Answer:

The time taken for the bullet to hit the ground: 1.1059

Final Velocity Vector: (200, 200, -10.85)

Final Position Vector: (221.18, 221.18, 0)

Explanation:

For this problem, lets start with finding the time it takes for the bullet to hit the ground. Since it was fired horizontally, it's initial vertical velocity is zero. This is also it's velocity in the z direction.

Time taken for the bullet to hit the ground:

$s=u*t +\frac{1}{2} (a*t^2)$

here, s = -6 m

u = 0 (initial velocity)

a = -9.81

Solving for t, we get 1.1059 seconds.

In this time, the position change in the x and y axis will be:

x axis: 200 * 1.1059 = 221.18

y axis: 200 * 1.1059 = 221.18

Thus the final position vector will be:

(221.18 , 221.18 , 0)

Note the zero in the z direction is because ground is located at 0.

Let's find the final z velocity as well for the velocity vector:

$v^2-u^2=2*a*s$

$v^2=2 * 9.81 * 6$

v = 10.85 m/s (final velocity)

The final velocity vector becomes:

(200, 200, -10.85)