A man shoots a rifle at a height of 1.7 m above level ground. His aim is perfectly horizontal. At the same time the bullet leaves the barrel of the gun at 670 m/s, the shell casing is ejected with a velocity of 3 m/s, also perfectly horizontal to the ground (but at an angle of 90° to the path of the bullet.)
Is there a question here?
I predict that the shell lands closer to the shooter than does the bullet!
Oh, did you want to ask a question, too?
Something tells me you will need to find out how long it takes an object to fall 1.7 meters.
I bet they both hit the ground at the same time, whenever that is.
To understand this scenario, we need to consider the horizontal motion of the bullet and the casing separately.
First, let's focus on the bullet:
The bullet is shot horizontally, which means it has no vertical acceleration. Therefore, its vertical position remains at a height of 1.7 m above the ground throughout its flight.
Since the bullet is fired with an initial velocity of 670 m/s, it will continue to move at a constant horizontal speed and will cover horizontal distance over time.
Now, let's consider the casing:
The casing is ejected with a velocity of 3 m/s in a horizontal direction. Since there is no vertical acceleration for the casing as well, its vertical position also remains at the same height of 1.7 m above the ground.
However, unlike the bullet, the casing is not moving with a constant speed because it is subject to the force of gravity. Over time, gravity will cause the casing to gradually fall toward the ground.
To calculate the horizontal distance traveled by the bullet and the casing, we need to know the time of flight. Since both the bullet and the casing are shot at the same time, they will impact the ground simultaneously. Therefore, we can determine the time of flight for both objects by using the equation of motion for vertical free fall:
h = (1/2) * g * t^2
In this equation, h represents the vertical displacement (1.7 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
Rearranging the equation, we find:
t = √(2h / g)
Substituting the given value of h, we get:
t = √(2 * 1.7 / 9.8) ≈ 0.615 seconds
Now, we can calculate the horizontal distance traveled by the bullet:
distance = horizontal velocity * time
distance = 670 m/s * 0.615 s ≈ 411.15 meters
Similarly, let's calculate the horizontal distance traveled by the casing:
distance = horizontal velocity * time
distance = 3 m/s * 0.615 s ≈ 1.845 meters
Therefore, the bullet will travel approximately 411.15 meters horizontally, while the casing will travel approximately 1.845 meters horizontally before hitting the ground.