A passenger riding on a train moving over a bridge with a velocity of 55m/s, drops a pen out of the window. the pen lands in the river below 2.4 seconds later.
A. find how high the window was above the water.
B. how far forward did the pen move?
To solve part A of the problem, we need to find the height of the window above the water when the pen was dropped.
First, let's write down the given values:
Initial velocity of the train, v = 55 m/s
Time taken for the pen to land, t = 2.4 seconds
Acceleration due to gravity, g = 9.8 m/s^2 (assuming no air resistance)
We can use the SUVAT (or kinematic) equations to solve this problem. The equation that relates displacement, initial velocity, time, and acceleration is:
s = ut + (1/2)at^2
In this case, the pen is dropped from rest (initial velocity, u, is zero). So the equation becomes:
s = (1/2)gt^2
Substituting the known values, we have:
s = (1/2) * 9.8 * (2.4)^2
Calculating this equation, we find that the height of the window above the water (s) is approximately 28.224 meters.
Therefore, the height of the window above the water is 28.224 meters.
Now let's move to part B of the problem, which asks for the distance the pen moved horizontally when it landed in the river.
Since the pen was dropped vertically, its horizontal motion will be unaffected by the train's velocity. Therefore, the distance the pen moved forward is simply the product of the train's velocity and the time taken for the pen to land.
Distance traveled horizontally = Velocity of the train * Time taken
Distance traveled horizontally = 55 m/s * 2.4 seconds
Calculating this equation, we find that the pen moved a distance of approximately 132 meters.
Therefore, the pen moved forward approximately 132 meters.