what is the total time and displacement if the traveler on a train throw ball on level track straight up relative to the train with a speed of 4.90m/s and velocity of 20.0m/s ?
Sorry its speed of 4.90m/s and the train velocity of 20.0m/s..
See my previous answer, earlier today, to what looks like the same question.
To determine the total time and displacement, we need to consider the motion of the ball as seen from the perspective of the train.
Let's break down the problem into two parts:
1. Upward motion of the ball:
When the traveler throws the ball upwards from the train, its initial velocity relative to the train is 4.90 m/s. The ball will travel upwards in an arc until it reaches its maximum height and then start coming back down due to gravity. At the highest point, its velocity will momentarily become 0 before it starts descending.
To calculate the total time for the upward motion, we can use the equation:
v = u + at
Where:
u = initial velocity (4.90 m/s, upwards)
v = final velocity (0 m/s, at the highest point)
a = acceleration (due to gravity, approximately -9.8 m/s^2, downwards)
t = total time
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 4.90) / (-9.8)
t ≈ 0.50 seconds
During this time, the displacement of the ball will be the vertical distance it has traveled upwards. We can calculate this using another equation:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (4.90 m/s, upwards)
t = total time (0.50 seconds)
a = acceleration (due to gravity, approximately -9.8 m/s^2, downwards)
Plugging in the values, we get:
s = (4.90 * 0.50) + (0.5 * -9.8 * (0.50)^2)
s ≈ 1.22 meters (upwards)
2. Downward motion of the ball:
After reaching the highest point, the ball starts coming back down with an initial velocity of 0 m/s. However, since the train is moving forward with a velocity of 20.0 m/s, the ball will also have a horizontal velocity relative to an observer on the ground.
The displacement in the horizontal direction is given by:
s = v * t
s = 20.0 * 0.50
s = 10.0 meters (horizontal)
The total displacement of the ball will be the vector sum of its displacement in the vertical and horizontal directions. This can be calculated using the Pythagorean theorem:
total displacement = √((vertical displacement)^2 + (horizontal displacement)^2)
total displacement = √((1.22)^2 + (10.0)^2)
total displacement ≈ 10.06 meters
Therefore, the total time taken by the ball and the displacement of the ball, relative to the ground, are approximately 0.50 seconds and 10.06 meters, respectively.