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
15.56m/s and 10.90m/s respectively
The vertical and horizontal components of a given vector, say A, are given by
[tex]A_{Y}[/tex] = A sin θ ----------------(i)
[tex]A_{X}[/tex] = A cos θ ----------------(ii)
Where;
[tex]A_{Y}[/tex] is the vertical component of the vector A
[tex]A_{X}[/tex] 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
[tex]V_{Y}[/tex] = V sin θ --------------------(iii)
[tex]V_{X}[/tex] = V cos θ ------------------------(iv)
Substitute all the necessary values into equations(iii) and (iv) as follows;
[tex]V_{Y}[/tex] = 19 sin 55° = 19m/s x 0.8192 = 15.56m/s
[tex]V_{X}[/tex] = 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.
