Upon reaching a speed of 215 km/h on the runway, a jet raises its nose to an

angle of 18º with the horizontal and begins to lift off the ground.

a. Calculate the horizontal and vertical components of its velocity
at this moment.

b. What is the physical interpretation of each of these components
of the jet’s velocity?

*how do I draw this properly? I drew a triangle and ended up with an angle of 162 across from the resultant. I solved for the other sides and got weird numbers. The textbook answer is a) 204km/h, 66km/h

horizontal speed: 215 cos18º

vertical: 215 sin18º

too bad you didn't show how you got your "weird' results...

To solve this problem, you can use trigonometry. Let's break it down step-by-step:

a. Calculating the horizontal and vertical components of the velocity:

1. Draw a diagram representing the situation. Label the horizontal velocity component as Vx and the vertical velocity component as Vy.

|
/ | \
/ | \
/ | \
/ θ | \
/ | \
-----------------
Vx
-----------------
Vy

2. Given that the magnitude of the velocity is 215 km/h, we can conclude that:

V = √(Vx^2 + Vy^2) ... (Equation 1)

3. We are also given that the angle between the velocity vector and the horizontal axis is 18º. Using this information, we can write the following equation:

tan(18º) = Vy / Vx ... (Equation 2)

4. Rearrange Equation 2 to solve for Vy:

Vy = Vx * tan(18º)

5. Substitute the value of Vy from Equation 4 into Equation 1:

V = √(Vx^2 + (Vx * tan(18º))^2)

6. Solve for Vx:

V^2 = Vx^2 + (Vx * tan(18º))^2
(215 km/h)^2 = Vx^2 + (Vx * tan(18º))^2

Solve this equation to find the value of Vx.

7. Once you find the value of Vx, substitute it back into Equation 4 to find the value of Vy.

b. Physical interpretation of the components:

The horizontal component (Vx) represents the speed of the jet in the horizontal direction. It tells us how fast the jet is moving parallel to the ground.

The vertical component (Vy) represents the speed of the jet in the vertical direction. It tells us how fast the jet is rising or descending vertically.

The angle θ (18º) represents the inclination of the jet's velocity vector with respect to the horizontal axis.

To solve this problem, you can use trigonometry to find the components of the jet's velocity.

Now, let's break down the problem step by step:

a. To calculate the horizontal and vertical components of the jet's velocity, use trigonometric functions, specifically sine and cosine. The given angle of 18º is with respect to the horizontal.

Step 1: Convert the speed in km/h to m/s.
Since we are dealing with trigonometric functions and SI units, it is better to work with meters per second (m/s) instead of kilometers per hour (km/h).

1 km/h = 1000 m / 3600 s ≈ 0.278 m/s

Hence, the speed of the jet is 215 km/h * 0.278 m/s = 59.87 m/s.

Step 2: Calculate the horizontal component of the velocity.
The horizontal component of the velocity (Vx) can be found using the cosine function:

Vx = speed * cos(angle)

Vx = 59.87 m/s * cos(18º)

Now, calculate Vx:

Vx = 59.87 m/s * cos(18º)
Vx = 59.87 m/s * 0.951
Vx ≈ 56.95 m/s

Therefore, the horizontal component of velocity is approximately 56.95 m/s.

Step 3: Calculate the vertical component of the velocity.
The vertical component of the velocity (Vy) can be found using the sine function:

Vy = speed * sin(angle)

Vy = 59.87 m/s * sin(18º)

Now, calculate Vy:

Vy = 59.87 m/s * sin(18º)
Vy = 59.87 m/s * 0.309
Vy ≈ 18.51 m/s

Therefore, the vertical component of velocity is approximately 18.51 m/s.

So, the horizontal component of the jet's velocity is approximately 56.95 m/s, and the vertical component is approximately 18.51 m/s.

b. The physical interpretation of each component of the jet's velocity is as follows:

- The horizontal component (Vx) represents the jet's speed in the direction parallel to the ground. It determines how fast the jet is traveling along the runway.

- The vertical component (Vy) represents the jet's speed in the direction perpendicular to the ground. The positive value indicates that the plane is ascending vertically.

Keep in mind that drawing a triangle can be helpful to visualize the problem. Here's a simple way to draw it:

1. Draw a horizontal line to represent the ground.
2. From one end of the line, draw a line slanting upwards at an angle of 18º. This represents the airplane's direction of ascent.
3. From the starting point, draw a vertical line upward to meet the slant line.
4. Label the horizontal line as "Vx" and the vertical line as "Vy."
5. Label the slanted line as "speed."

Now you have a triangle with the speed as the hypotenuse, and the vertical and horizontal lines as its legs.