Question:

A vector makes an angle of 30^o with the horizontal. If the horizontal component of the vector is 250.Find the magnitude of the vector and its vertical component.

Solution: Vector magnitude and vertical component:

Horizontal component of the vector is 30 degree (Given)

$${A_x=A \cos\theta}$$

Vertical component of the vector

$${A_y=A \sin\theta}$$

Magnitude of the vector is A

The angle in which the vector makes with the horizontal is 30 degree.

To find the magnitude of the vector:

The magnitude of the vector is calculated from the horizontal component value A_x

$${A_x =A \cos\theta}$$

$${250=A \cos 30^\circ}$$

$${250 =A \frac{\sqrt3}{2}}$$

$${A=\frac{500}{1.732}}$$

$${A=289}$$

The magnitude of the vector A is approximately 289

To find the vertical component A_y of the vector:

$${A_y=A \sin \theta}$$

$${A_y=289 \ \sin30^\circ}$$

$${A_y = (289) (\frac{1}{2})}$$

$${A_y=144.5}$$

The vertical component of the vector V_y is 144.5 (Approximately)

Conclusion:

Horizontal component of the vector A_x is 250(Given).

The angle in which the vector makes with the horizontal is 30 degree(Given).

The magnitude of the vector A is approximately 289

The vertical component of the vector A_y is 144.5 (Approximately)

Vector magnitude and vertical component- Practical Examples

Example 1:

The Takeoff of an Aircraft:

At a 30° angle to the ground, an aeroplane ascends. What is its true speed (magnitude) and rate of vertical ascent if its horizontal speed (ground speed) is 250 km/h?

Solution:

Given:

30 degrees is the angle 𝛳.

250 km/h is the horizontal component Vx.

Determine the velocity V’s magnitude.

The plane’s actual speed is about 288.68 km/h.

Find the vertical component Vy

The plane is climbing at 144.34 km/h

Interpretation:

Even if the plane moves 250 km/h forward, its overall speed is higher (288.68 km/h) because it’s also climbing. Such computations are used by air traffic controllers to control climb rates.

Example 2: A Missile Launch

To strike a far-off target, a missile is fired at a 30° angle. How fast does it launch and rise vertically if its horizontal velocity component is 250 m/s?

Solution:

Given:

Angle 𝛳 = 30 degrees

Vx, the horizontal component, equals 250 m/s.

Determine the velocity V’s magnitude.

The launch velocity of the missile is 288.68 m/s.

Locate the vertical component, Vy.

144.34 m/s is the missile’s rising speed.

Interpretation:  

In order to compensate for the upward motion, the missile must move faster than its horizontal speed. This is essential for figuring out impact time and range.

Example 3: A Ramp for Loading Goods

In order to load packages onto a truck, a conveyor ramp is angled at a thirty-degree angle. What is the conveyor belt’s actual speed and the vertical lifting speed if the items are moving horizontally at 250 cm/s?

Solution:

Given:

Angle 𝛳 = 30 degrees

Component horizontal Vx = 250 cm/s

Determine the velocity V’s magnitude.

The speed of the conveyor belt is 288.68 cm/s.

Locate the vertical component of Vy.

Packages are raised at a speed of 144.34 cm/s.

Interpretation:

For packages to go upward efficiently, the belt’s speed must be greater than the horizontal speed.