Question

a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical and horizontal velocity components.

Answer 1

Answer:

15.56m/s and 10.90m/s respectively

Explanation:

The vertical and horizontal components of a given vector, say A, are given by

$A_{Y}$ = A sin θ                  ----------------(i)

$A_{X}$ = A cos θ                 ----------------(ii)

Where;

$A_{Y}$ is the vertical component of the vector A

$A_{X}$ is the horizontal component of the vector A

A is the magnitude of the vector A

θ is the angle the vector makes with the positive x-axis (horizontal direction).

Now, from the question;

The vector is the velocity of the 1-kg discus. Lets call it vector V

The magnitude of the velocity vector V = V = 19m/s

The angle that the vector makes with the positive x-axis = θ

To calculate θ;

Notice that the velocity vector makes an angle of 35° from the vertical direction rather than the horizontal direction.

Therefore, to get the horizontal direction of the velocity vector, we subtract 35° from 90° as follows;

θ = 90° - 35° = 55°

Now, the vertical and horizontal components of the velocity vector, V, are given by

$V_{Y}$ = V sin θ              --------------------(iii)

$V_{X}$ = V cos θ             ------------------------(iv)

Substitute all the necessary values into equations(iii) and (iv) as follows;

$V_{Y}$ = 19 sin 55° = 19m/s x 0.8192 = 15.56m/s

$V_{X}$ = 19 cos 55° = 19m/s x 0.5736 = 10.90m/s

Therefore, the vertical and horizontal velocity components are respectively 15.56m/s and 10.90m/s.

Answer 2

Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s  × cos 55°

Vx =  10.9 m/s

Vx = V sin θ

Vy = 19 m/s  × sin 55°

Vy = 15.6 m/s