a car travels due east with a speed of 42.0 km/h. raindrops are falling at a constant speed vertically with…

Explanation:

Step1: Analyze velocity components

Let the velocity of the car be $v_{c}=42.0$ km/h and the angle of the rain - trace on the side - window with the vertical be $\theta = 64.0^{\circ}$. Let the velocity of the rain with respect to the Earth be $v_{rE}$ (vertical) and the velocity of the rain with respect to the car be $v_{rc}$. The horizontal component of the velocity of the rain with respect to the car is equal to the velocity of the car (due to relative motion), i.e., $v_{rcx}=v_{c}$ and the vertical component is $v_{rcy}=v_{rE}$.

Step2: Find the velocity of the rain with respect to the car

We know that $\tan\theta=\frac{v_{rcx}}{v_{rcy}}$. Since $v_{rcx} = v_{c}$ and we want to find $v_{rc}$, and using the right - triangle relationship for the velocity of the rain with respect to the car $v_{rc}=\frac{v_{c}}{\sin\theta}$. Substitute $v_{c} = 42.0$ km/h and $\theta = 64.0^{\circ}$ into the formula: $v_{rc}=\frac{42.0}{\sin64.0^{\circ}}=\frac{42.0}{0.8988}\approx46.7$ km/h.

Step3: Find the velocity of the rain with respect to the Earth

Since $\tan\theta=\frac{v_{c}}{v_{rE}}$, then $v_{rE}=\frac{v_{c}}{\tan\theta}$. Substitute $v_{c} = 42.0$ km/h and $\theta = 64.0^{\circ}$ into the formula: $v_{rE}=\frac{42.0}{\tan64.0^{\circ}}=\frac{42.0}{2.0503}\approx20.5$ km/h.

Answer:

(a) 46.7 km/h (b) 20.5 km/h