Two boats are heading away from shore. Boat 1 heads due north at a speed of 4 m/s relative to the shore. Relative to boat 1, boat 2 is moving 35° north of east at a speed of 1.2 m/s. A passenger on boat 2 walks due east across the deck at a speed of 1.3 m/s relative to boat 2. What is the speed of the passenger relative to the shore?

North:

4 + 1.2 sin 35

East
1.2 cos 35 + 1.3

An arrow is short into the air with an initial Velocity of 100m\s at an elevation of 6 degrees find (a)the time of lilght (b) the maximum lenght attained (c) the range

To find the speed of the passenger relative to the shore, we need to consider the velocities of boat 1, boat 2, and the passenger.

Let's break down the velocities:

1. Boat 1 is moving due north at a speed of 4 m/s relative to the shore.
2. Boat 2 is moving 35° north of east at a speed of 1.2 m/s relative to boat 1.
3. The passenger on boat 2 is walking due east across the deck at a speed of 1.3 m/s relative to boat 2.

To find the speed of the passenger relative to the shore, we can use vector addition.

First, let's break down the velocity of boat 2 into its eastward and northward components relative to the shore:

Eastward component: 1.2 m/s * cos(35°)
Northward component: 1.2 m/s * sin(35°)

Next, let's determine the velocity of the passenger relative to the shore.

The passenger's velocity relative to the shore will be the sum of their velocity relative to boat 2 and boat 2's velocity relative to the shore.

The passenger's velocity relative to boat 2 is 1.3 m/s in the eastward direction.

Adding the eastward components together:
Eastward component: 1.2 m/s * cos(35°) + 1.3 m/s

Adding the northward components together:
Northward component: 1.2 m/s * sin(35°)

Now, we can use the Pythagorean theorem to find the magnitude (speed) of the passenger's velocity relative to the shore:

Magnitude of velocity = √[(Eastward component)^2 + (Northward component)^2]

Magnitude of velocity = √[(1.2 m/s * cos(35°) + 1.3 m/s)^2 + (1.2 m/s * sin(35°))^2]

Calculating this expression will give us the speed of the passenger relative to the shore.

To find the speed of the passenger relative to the shore, we need to calculate the vector sum of the velocities of boat 2 and the passenger relative to the shore.

Let's break down the problem into components:

The velocity of boat 1 is directly north, so its velocity components are (0, 4 m/s) relative to the shore.

The velocity of boat 2 is 35° north of east and has a magnitude of 1.2 m/s. To find its velocity components relative to the shore, we can use trigonometry. The horizontal component is given by cos(35°) * 1.2 m/s, and the vertical component is given by sin(35°) * 1.2 m/s.

The velocity of the passenger is directly east relative to boat 2, so its velocity components relative to boat 2 are (1.3 m/s, 0).

Now, let's add up the velocities to calculate the velocity of the passenger relative to the shore:

The x-component of the passenger's velocity relative to the shore is the sum of the x-components of boat 2's velocity (cos(35°) * 1.2 m/s) and the passenger's velocity (1.3 m/s). So, the x-component is equal to cos(35°) * 1.2 m/s + 1.3 m/s.

The y-component of the passenger's velocity relative to the shore is the sum of the y-components of boat 2's velocity (sin(35°) * 1.2 m/s) and the passenger's velocity (0). So, the y-component is equal to sin(35°) * 1.2 m/s + 0.

Using these components, we can use the Pythagorean theorem to find the magnitude of the passenger's velocity relative to the shore:

Magnitude = √(x-component^2 + y-component^2)

Finally, substitute the values into the equation and calculate the magnitude of the passenger's velocity relative to the shore.