Question
A car travels due east with a speed of 35.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth. The traces of the rain on the side windows of the car make an angle of 40.0° with the vertical. Find the velocity of the rain with respect to the following reference frames.
a) car
b) earth
a) car
b) earth
Answers
a) They have told you that the rain is falling vertically. Call its vertical velocity Vy. In the reference frame of the car, the rain also has a horizontal component of Vx = 35 km/h and a vertical component Vy. Because of the 40 degree tracks of the drops,
Vx/Vy = tan 40 = 0.839
Vy = 1.19*Vx = 41.7 km/h
Magnitude of velocity = sqrt (Vx^2 + Vy^2) = 54.5 km/h
b) In the earth references frame, the only velocity componenet is the vertical one, Vx, which is the same as in part (a)
Vx/Vy = tan 40 = 0.839
Vy = 1.19*Vx = 41.7 km/h
Magnitude of velocity = sqrt (Vx^2 + Vy^2) = 54.5 km/h
b) In the earth references frame, the only velocity componenet is the vertical one, Vx, which is the same as in part (a)
56
car: 35km/h divided by tan (40) = 41.7 sqrt of 35^2+41.7^2=54.4
earth: use Pythagorean theorem sqrt 54.4^2-35^2=41.6
use the same 40 degrees in the answer
Unsure if earth is right, but I had 38.0km/h at 62 degrees for my problem and used those steps to get 43.0 km/h for the car and 20.1km/h for Earth. They were correct so here is my input
earth: use Pythagorean theorem sqrt 54.4^2-35^2=41.6
use the same 40 degrees in the answer
Unsure if earth is right, but I had 38.0km/h at 62 degrees for my problem and used those steps to get 43.0 km/h for the car and 20.1km/h for Earth. They were correct so here is my input