A passenger on a moving train tosses a coin with an initial velocity of (-2.25 m/s)x unit vector + 4.26 m/s)y unit vector (from the point of view of the passenger). The train's velocity relative to the ground is (11.5 m/s)x unit vector.
(a) What is the minimum speed of the coin relative to the ground during its flight?
At what part in the coin's flight does this minimum speed occur?
choose one from the following
1. immediately after being tossed
2. at some point as it is rising
3. at the top of its motion
4. at some point at it is falling
(b) Find the initial speed and direction of the coin as seen by an observer on the ground.
( ) m/s
( ) degrees (counterclockwise from the +x axis
(c) Use the expression for ymax derived in Example 4-7 to calculate the maximum height of the coin, as seen by an observer on the ground.
( )m
(d) Repeat part (c) from the point of view of the passenger.
( )m
Please help! I don't even know where to start!
To solve this problem, you can break it down into several steps. Let's go through each step:
Step 1: Decompose the velocities
The first step is to decompose the velocities of the coin and the train into their x and y components. From the given information:
Coin's velocity (from the passenger's point of view):
V_coin_x = -2.25 m/s
V_coin_y = 4.26 m/s
Train's velocity (relative to the ground):
V_train_x = 11.5 m/s
Step 2: Find the minimum speed of the coin relative to the ground
To find the minimum speed of the coin relative to the ground, we need to combine the velocities of the coin and the train.
V_coin_x_relative = V_coin_x + V_train_x
V_coin_y_relative = V_coin_y
Now, we can find the magnitude of the velocity vector:
V_coin_relative = sqrt(V_coin_x_relative^2 + V_coin_y_relative^2)
Step 3: Determine the part of the flight where the minimum speed occurs
To determine where the minimum speed occurs, we need to consider the vertical motion of the coin. The minimum speed occurs when the coin reaches its maximum height, which is at the top of its motion (option 3).
Step 4: Find the initial speed and direction of the coin as seen by an observer on the ground
To find the initial speed and direction of the coin as seen by an observer on the ground, we need to find the magnitude and direction of the absolute velocity vector.
V_coin_absolute = sqrt(V_coin_x^2 + V_coin_y^2)
θ_coin_absolute = atan(V_coin_y / V_coin_x)
Step 5: Calculate the maximum height of the coin as seen by an observer on the ground
We can use the expression for maximum height derived in Example 4-7:
y_max = (V_coin_relative_y^2) / (2 * g)
Step 6: Calculate the maximum height of the coin as seen by the passenger
To calculate the maximum height of the coin as seen by the passenger, we need to consider the motion relative to the passenger.
y_max_passenger = (V_coin_y^2) / (2 * g)
Now that we have gone through the steps, let's calculate the missing values:
(a) The minimum speed of the coin relative to the ground occurs at the top of its motion (option 3). You can calculate it using the values obtained in Step 2.
(b) To find the initial speed and direction of the coin as seen by an observer on the ground, use the values obtained in Step 4.
(c) To calculate the maximum height of the coin as seen by an observer on the ground, use the value obtained in Step 5.
(d) To calculate the maximum height of the coin as seen by the passenger, use the value obtained in Step 6.
Once you have calculated the values, you can fill them in the respective blanks. Remember to use the correct units and round your answers appropriately.