Question

if a vehicle is moving at a velocity u, and theres a gun attached to its back, and a bullet shoots out opposite to the direction of motion of the vehicle, with a velocity v at an angle of 45 degrees with respect to the horizontal, what is the distance between the vehicle and bullet when the bullet falls?

Answer 1

v^2/g

Answer 2

To determine the distance between the vehicle and the bullet when the bullet falls, we need to analyze the motion of both the vehicle and the bullet.

Let's set up a coordinate system where the vehicle is moving in the positive x-direction.
Let the initial position of the gun (at the back of the vehicle) be the origin (0,0) at time $t=0$.

1. Velocity of the vehicle:
   The vehicle moves at a constant velocity $u$ in the positive x-direction.
   $V_{vehicle, x} = u$

2. Velocity of the bullet relative to the gun (vehicle):
   The bullet is shot with a velocity $v$ at an angle of 45 degrees with respect to the horizontal, opposite to the direction of motion of the vehicle.
   This means the horizontal component of the bullet's velocity relative to the gun is in the negative x-direction.
   $v_{bullet/gun, x} = -v \cos(45^\circ) = -v \frac{1}{\sqrt{2}}$
   $v_{bullet/gun, y} = v \sin(45^\circ) = v \frac{1}{\sqrt{2}}$

3. Velocity of the bullet relative to the ground:
   The velocity of the bullet relative to the ground is the vector sum of its velocity relative to the gun and the gun's velocity relative to the ground (which is the vehicle's velocity).
   $V_{bullet, x} = v_{bullet/gun, x} + V_{vehicle, x} = -v \frac{1}{\sqrt{2}} + u$
   $V_{bullet, y} = v_{bullet/gun, y} + 0 = v \frac{1}{\sqrt{2}}$ (since the vehicle has no vertical velocity)

4. Time of flight of the bullet:
   The bullet "falls" when its vertical displacement is zero (assuming it starts and ends at the same height). We use the vertical motion equation:
   $y = V_{bullet, y} t - \frac{1}{2}gt^2$
   Setting $y=0$ for the time of flight $t$:
   $0 = \left(v \frac{1}{\sqrt{2}}\right)t - \frac{1}{2}gt^2$
   $t \left(v \frac{1}{\sqrt{2}} - \frac{1}{2}gt\right) = 0$
   The non-zero solution for $t$ is:
   $v \frac{1}{\sqrt{2}} = \frac{1}{2}gt$
   $t = \frac{2v}{g\sqrt{2}} = \frac{v\sqrt{2}}{g}$

5. Distance between the vehicle and the bullet:
   The distance between the vehicle and the bullet is the magnitude of their relative horizontal displacement. This can be calculated by multiplying the relative horizontal velocity by the time of flight.
   The horizontal velocity of the bullet relative to the vehicle is:
   $V_{bullet/vehicle, x} = V_{bullet, x} - V_{vehicle, x}$
   $V_{bullet/vehicle, x} = \left(u - v \frac{1}{\sqrt{2}}\right) - u = -v \frac{1}{\sqrt{2}}$
   The distance between them is the magnitude of this relative velocity multiplied by the time of flight:
   Distance $D = |V_{bullet/vehicle, x}| \times t$
   $D = \left|-v \frac{1}{\sqrt{2}}\right| \times \left(\frac{v\sqrt{2}}{g}\right)$
   $D = \left(v \frac{1}{\sqrt{2}}\right) \times \left(\frac{v\sqrt{2}}{g}\right)$
   $D = \frac{v^2 \times 2}{2g}$
   $D = \frac{v^2}{g}$

The distance between the vehicle and the bullet when the bullet falls is $\frac{v^2}{g}$. This result is independent of the vehicle's speed $u$.

The final answer is $\frac{v^2}{g}$.

Explanation of the solution: The time of flight of the bullet is determined by its vertical velocity component relative to the ground, which is $v \sin(45^\circ) = v/\sqrt{2}$. Using the projectile motion formula for time of flight ($T = 2V_y/g$), we get $T = (2 \cdot v/\sqrt{2})/g = v\sqrt{2}/g$. The horizontal velocity of the bullet relative to the vehicle is its horizontal component of velocity relative to the gun, which is $-v \cos(45^\circ) = -v/\sqrt{2}$ (negative because it's fired opposite to the vehicle's motion). The distance between the vehicle and the bullet is the product of this relative horizontal speed and the time of flight: Distance $= |-v/\sqrt{2}| \times (v\sqrt{2}/g) = (v/\sqrt{2}) \times (v\sqrt{2}/g) = v^2/g$.