bullet traveling 8.o*10^2 (800) meters per second horizontally hits a target 180 m away. how far does the bullet fall before it hits the target?
To determine how far the bullet falls before hitting the target, we will use the equations of motion for projectile motion.
First, let's break down the problem into its components:
- Initial velocity in the horizontal direction (Vx) = 800 m/s
- Vertical acceleration (ay) = -9.8 m/s^2 (assuming near the Earth's surface)
- Distance traveled horizontally (range) = 180 m
We know that the bullet's vertical motion is affected by the force of gravity. The bullet's vertical velocity (Vy) starts at 0 m/s and continuously increases downward due to the acceleration caused by gravity.
The time taken for the bullet to reach the target can be found using the equation of motion for horizontal motion:
range = Vx * t
180 m = 800 m/s * t
t = 0.225 s
Now that we have the time taken for the bullet to reach the target, we can calculate the vertical displacement (distance fallen) using the equation of motion for vertical motion:
displacement = (1/2) * ay * t^2
displacement = (1/2) * (-9.8 m/s^2) * (0.225 s)^2
displacement = -0.125 m (rounded to three decimal places)
Therefore, the bullet falls approximately 0.125 meters (or 12.5 cm) before hitting the target.