what is the vert. comp. of vector with magnitude 88.5m/s and whose horiz. comp. is 75.4m/s? what is the direct. of this vector?
cos theta = 75.4/88.5
theta = cos^-1 .852
theta = 31.57 degrees from horizonal
vertical component = 88.5 sin 31.57
=46.33
thanks, i figured it out
To find the vertical component of the vector, we can use the Pythagorean theorem. The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two other sides. In this case, the hypotenuse represents the magnitude of the vector, the horizontal component represents one side, and the vertical component represents the other side.
Let's denote the vertical component as Vy and the horizontal component as Vx. We have the following information:
Magnitude of the vector (hypotenuse) = 88.5 m/s
Horizontal component (Vx) = 75.4 m/s
Using the Pythagorean theorem, we can calculate the vertical component as follows:
Vy^2 + Vx^2 = Magnitude^2
Vy^2 + (75.4 m/s)^2 = (88.5 m/s)^2
Simplifying the equation, we have:
Vy^2 + 5679.16 m^2/s^2 = 7822.25 m^2/s^2
Subtracting 5679.16 m^2/s^2 from both sides of the equation, we get:
Vy^2 = 7822.25 m^2/s^2 - 5679.16 m^2/s^2
Vy^2 = 2143.09 m^2/s^2
Taking the square root of both sides, we find:
Vy ≈ √2143.09 m^2/s^2
Vy ≈ 46.28 m/s
Therefore, the vertical component of the vector is approximately 46.28 m/s.
To find the direction of the vector, we can use trigonometry. We can use the inverse tangent (arctan) function to find the angle of the vector with respect to the horizontal axis.
The formula for this is: θ = arctan(Vy / Vx)
Plugging in the values, we get:
θ = arctan(46.28 m/s / 75.4 m/s)
Using a calculator or trigonometric table, we find:
θ ≈ 31.5 degrees
Therefore, the direction of the vector is approximately 31.5 degrees above the horizontal axis.