You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -40.0 units. What are the two possibilities for its x component?
what is arcsin( 40/90)?
Then x is +- 90Cos above angle
To find the two possibilities for the x component of the vector, we can use the magnitude and the y component. Let's denote the x component as 'x' and the y component as 'y'.
The magnitude of the vector can be calculated using the Pythagorean theorem:
magnitude = sqrt(x^2 + y^2)
Given that the magnitude is 90.0 units and the y component is -40.0 units, we can substitute these values into the equation:
90.0 = sqrt(x^2 + (-40.0)^2)
Solving for x:
90.0 = sqrt(x^2 + 1600)
Squaring both sides of the equation:
8100.0 = x^2 + 1600
Rearranging the equation:
x^2 = 8100.0 - 1600
x^2 = 6500.0
Taking the square root of both sides:
x = ±sqrt(6500.0)
The two possibilities for the x component are the positive and negative square root of 6500.0:
x = ±80.62
Therefore, the two possibilities for the x component are +80.62 and -80.62 units.