A man walking on straight path at15km/h finds that the rain is falling at an angle of 45degree to the vertical.As the man increases his speed to 20km/h,he finds the rain now falling at 30degree with the vertical. What is the true velocity of the rain.

I don't follow the statement. How can what the man does, change the path of the rain? What if the man stops? or goes inside?

Or, or you moving the vertical frame of reference?

man walking towards left..(¡û)

To find the true velocity of the rain, we need to break down the problem using vector addition.

Let's consider two vectors:
1. The velocity of the man relative to the ground, denoted as Vm.
2. The velocity of the rain relative to the ground, denoted as Vr.

When the man is walking at 15 km/h, the rain appears to fall at a 45-degree angle to the vertical. This means that the vertical component of the rain's velocity is equal to the man's velocity, Vm.

Now, when the man increases his speed to 20 km/h, the rain appears to fall at a 30-degree angle to the vertical. Let's denote the angle between Vr and the vertical as θ. In this case, the vertical component of Vr can be calculated as Vr * cos(θ).

We know that the horizontal component of the rain's velocity remains the same (since it's not affected by the man's motion), so let's denote it as Vr(H).

Now, we can set up the equations:

Equation 1: Vm = Vr (since the vertical component of the rain's velocity appears to be the same as the man's velocity)

Equation 2: Vm = Vr * cos(θ) (since the vertical component of the rain's velocity is Vr * cos(θ))

From equation 1, we can substitute Vr with Vm in equation 2:

Vm = Vm * cos(θ)

Now, we solve for cos(θ):

cos(θ) = 1

Since cos(θ) = 1, this means that θ = 0 degrees, which implies that the vertical component of the rain's velocity is equal to the rain's overall velocity.

Therefore, the true velocity of the rain is equal to the man's velocity, which is 15 km/h.