Question

a plane is flying at 115 m/s in the y-direction. the wind is blowing at 75.0 m/s in the x-direction. what is the direction of the velocity of the plane? \\( \theta = ? ^\circ \\)

Explanation:

Step1: Identify velocity components

The plane's velocity in y - direction, \(v_y = 115\space m/s\), and wind (which affects the plane's velocity in x - direction) \(v_x=75.0\space m/s\). To find the direction \(\theta\) (with respect to x - axis or y - axis? Wait, since \(v_y\) is along y and \(v_x\) along x, the angle \(\theta\) from the x - axis (or we can calculate the angle whose tangent is \(\frac{v_y}{v_x}\) if we consider the angle above x - axis) or wait, actually, the velocity of the plane relative to the ground is the vector sum of its velocity relative to air and wind velocity. So \(v_x = 75\space m/s\), \(v_y=115\space m/s\). The direction \(\theta\) (measured from the x - axis towards y - axis) is given by \(\tan\theta=\frac{v_y}{v_x}\)

Step2: Calculate the tangent of the angle

\(\tan\theta=\frac{115}{75.0}\)
\(\tan\theta=\frac{23}{15}\approx1.5333\)

Step3: Find the angle

\(\theta=\arctan(1.5333)\)
Using a calculator, \(\arctan(1.5333)\approx56.9^{\circ}\) (we can also check: \(\tan(56.9^{\circ})\approx1.53\), which is close to \(\frac{23}{15}\approx1.533\))

Answer:

\(56.9\) (or more precisely, if we calculate \(\arctan(\frac{115}{75})\): \(\frac{115}{75}=\frac{23}{15}\approx1.5333\), \(\arctan(1.5333)\approx56.9^{\circ}\))