An object is shot from the back of a railroad flatcar moving at 40 km/h on a straight horizontal
road. The launcher is aimed upward, perpendicular to the bed of the flatcar. The object falls:
a) in front of the flatcar
b) behind the flatcar
c) on the truck
d) either behind or in front of the flatcar, depending on the initial speed of the object
e) to the side of the flatcar
On the flatcar
The object will fall in front of the flatcar.
This can be understood by considering the motion of the object relative to the flatcar. When the object is shot from the launcher, it will inherit the velocity of the flatcar, which is 40 km/h in the forward direction.
Since the launcher is aimed upward perpendicular to the bed of the flatcar, the initial velocity of the object will have both a horizontal component (due to the velocity of the flatcar) and a vertical component (due to the aim of the launcher). However, the horizontal component of the object's velocity is the same as the flatcar's velocity, so it will not change during the object's motion.
The object will experience only vertical acceleration due to gravity, and there is no horizontal force acting on it. As a result, the horizontal component of the object's velocity will remain constant throughout its fall. Thus, the object will fall in front of the flatcar along the same straight horizontal path, in the direction of the initial horizontal velocity.
To determine where the object falls, we need to consider the horizontal velocity of the object and the vertical motion separately.
1. Horizontal motion:
Since the object is shot from the back of the flatcar, it will initially have the same horizontal velocity as the flatcar. Therefore, the horizontal velocity of the object is 40 km/h.
2. Vertical motion:
The vertical motion of the object is governed by gravity. As there is no horizontal force acting on the object once it is in the air, its horizontal velocity remains constant throughout its trajectory. Only gravity acts on the object vertically, causing it to accelerate downwards.
Considering the vertical motion, we need to determine the time it takes for the object to fall. The time of flight can be found using the equation for vertical motion:
h = (1/2) * g * t^2,
where h is the vertical distance traveled, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
Since we don't have the value of the vertical distance, we will assume the object falls from a height close to the flatcar's height. This allows us to ignore the effect of air resistance.
Now, substituting the values:
h = 0.5 * 9.8 * t^2,
h = 4.9 * t^2.
Next, we need to find the value of t when the object hits the ground. During this time, the flatcar and the object will continue to move horizontally.
t = (2 * h / g)^0.5,
t = (2 * 1 / 9.8)^0.5 = 0.451 s (approx).
Finally, we can determine the horizontal distance traveled by the flatcar and the object during this time. Using the equation:
d = v * t,
where d is the horizontal distance, v is the velocity, and t is the time, we can calculate:
d = 40 km/h * 0.451 s = 18.04 km (approx).
Therefore, the object will fall behind the flatcar, option (b), as it travels a shorter horizontal distance than the flatcar during the time it takes to fall.