A passenger on a moving train tosses a coin with an initial velocity of (-2.25 m/s) + 4.68 m/s) (from the point of view of the passenger). The train's velocity relative to the ground is (11.5 m/s).

(a) What is the minimum speed of the coin relative to the ground during its flight?
______m/s

(b) At what part in the coin's flight does this minimum speed occur?
-immediately after being tossed
-at some point as it is rising
-at the top of its motion
-at some point as it is falling
Explain.

(c)Find the initial speed and direction of the coin as seen by an observer on the ground.
____m/s
____° (counterclockwise from the +x axis)
(d) 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

(e) Repeat part (d) from the point of view of the passenger.
____m

(a) To find the minimum speed of the coin relative to the ground during its flight, we need to analyze the velocity of the coin relative to the ground.

The velocity of the coin relative to the ground can be found by adding the velocity of the train to the velocity of the coin relative to the train.

Given:
Velocity of the coin relative to the train (vct) = (-2.25 m/s, 4.68 m/s) relative to the passenger
Velocity of the train relative to the ground (vtg) = (11.5 m/s) in the same direction of the train's motion

To find the velocity of the coin relative to the ground (vcg), we add the velocities vectorially:
vcg = vct + vtg
vcg = (-2.25 m/s, 4.68 m/s) + (11.5 m/s, 0)
vcg = (9.25 m/s, 4.68 m/s)

The minimum speed of the coin relative to the ground is the magnitude of vcg, which can be found using the Pythagorean theorem:
Min Speed = sqrt((9.25 m/s)^2 + (4.68 m/s)^2)
Min Speed = sqrt(85.5625 + 21.9024)
Min Speed = sqrt(107.4649)
Min Speed ≈ 10.37 m/s

Therefore, the minimum speed of the coin relative to the ground during its flight is approximately 10.37 m/s.

(b) The minimum speed occurs at the top of the coin's flight. This is because at the top, the vertical velocity component approaches zero before changing direction. The coin momentarily stops moving upwards and then starts moving downwards. At this point, the speed of the coin relative to the ground reaches its minimum.

Therefore, the minimum speed occurs at the top of the coin's motion.

(c) 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 angle of the velocity vector vcg.

The magnitude of vcg is the speed of the coin relative to the ground, which we found in part (a) as approximately 10.37 m/s.

The direction of vcg can be found by calculating the angle it makes with the positive x-axis. We can use trigonometry to do this. Let θ be the angle between vcg and the x-axis.

θ = arctan(vcg_y / vcg_x)
θ = arctan(4.68 m/s / 9.25 m/s)
θ ≈ 26.21°

Therefore, the initial speed of the coin as seen by an observer on the ground is approximately 10.37 m/s, and the angle counterclockwise from the +x axis is approximately 26.21°.

(d) To calculate the maximum height of the coin as seen by an observer on the ground, we can use the formula derived in Example 4-7.

The formula for the maximum height (ymax) is given by:
ymax = (v0^2 * sin^2(θ)) / (2 * g)

Given:
v0 = initial speed of the coin relative to the ground (10.37 m/s)
θ = launch angle relative to the ground (26.21°)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Substituting these values into the formula:
ymax = (10.37^2 * sin^2(26.21°)) / (2 * 9.8)
ymax = (107.4649 * 0.21661443) / 19.6
ymax ≈ 1.179 m

Therefore, the maximum height of the coin as seen by an observer on the ground is approximately 1.179 m.

(e) To calculate the maximum height of the coin as seen by the passenger, we need to consider the launch angle from the passenger's perspective.

The launch angle from the passenger's perspective is the same as the launch angle relative to the ground (26.21°), as they are both measured with respect to the horizontal direction.

Using the same formula as in part (d):
ymax = (v0^2 * sin^2(θ)) / (2 * g)

Substituting the given values:
ymax = (10.37^2 * sin^2(26.21°)) / (2 * 9.8)
ymax = (107.4649 * 0.21661443) / 19.6
ymax ≈ 1.179 m

Therefore, the maximum height of the coin as seen by the passenger is also approximately 1.179 m.