A girl standing on road holds her umbrella at 45° with the vertical to keep the rain away. If she starts running without umbrella with a speed of \(15\sqrt2\) \(kmh^{–1}\), the rain drops hit herhead vertically. The speed of rain drops with respect to the moving girl is

Question

Which of the following is NOT a layer of the urinary bladder?

Answer 1

To solve this problem, let's start by analyzing the situation and using the laws of relative motion.

When the girl is standing still with her umbrella tilted at \(45^\circ\) with respect to the vertical, the rain is hitting her umbrella at this angle. This angle indicates the component of rain's horizontal velocity is equal to its vertical component.
When she starts running with a velocity of \(15\sqrt{2} \, \text{kmh}^{-1}\), the rain now falls vertically from her perspective. This means that the horizontal component of the rain's velocity equals the girl's running speed to counterbalance it.
Since the rain drops appear vertical to the moving girl, the rain's horizontal velocity (\(v_{\text{rain, horizontal}}\)) equals the girl's velocity: v_{\text{rain, horizontal}} = 15\sqrt{2} \, \text{kmh}^{-1}.
Initially, while stationary, the horizontal and vertical components of the rain's velocity are equal \(\text{(because of the 45° angle)}\): v_{\text{rain, vertical}} = v_{\text{rain, horizontal}}.
Therefore, the rain's vertical component is also \(15\sqrt{2} \, \text{kmh}^{-1}\).
To find the speed of the rain drops relative to the moving girl, we only consider the vertical component because the horizontal component is canceled by her motion:
The vertical component alone is: v_{\text{rain, relative}} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{2 \times (15\sqrt{2})^2} = \sqrt{2 \times 450}\ = 30 \, \text{kmh}^{-1}\.
Hence, the speed of the rain drops with respect to the moving girl is: \frac{30}{\sqrt{2}} \, \text{kmh}^{-1}\.

Therefore, the correct answer is \frac{30}{\sqrt{2}} \, \text{kmh}^{-1}\.

Answer 2

To solve this problem, let's start by analyzing the situation and using the laws of relative motion.

When the girl is standing still with her umbrella tilted at \(45^\circ\) with respect to the vertical, the rain is hitting her umbrella at this angle. This angle indicates the component of rain's horizontal velocity is equal to its vertical component.
When she starts running with a velocity of \(15\sqrt{2} \, \text{kmh}^{-1}\), the rain now falls vertically from her perspective. This means that the horizontal component of the rain's velocity equals the girl's running speed to counterbalance it.
Since the rain drops appear vertical to the moving girl, the rain's horizontal velocity (\(v_{\text{rain, horizontal}}\)) equals the girl's velocity: v_{\text{rain, horizontal}} = 15\sqrt{2} \, \text{kmh}^{-1}.
Initially, while stationary, the horizontal and vertical components of the rain's velocity are equal \(\text{(because of the 45° angle)}\): v_{\text{rain, vertical}} = v_{\text{rain, horizontal}}.
Therefore, the rain's vertical component is also \(15\sqrt{2} \, \text{kmh}^{-1}\).
To find the speed of the rain drops relative to the moving girl, we only consider the vertical component because the horizontal component is canceled by her motion:
The vertical component alone is: v_{\text{rain, relative}} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{2 \times (15\sqrt{2})^2} = \sqrt{2 \times 450}\ = 30 \, \text{kmh}^{-1}\.
Hence, the speed of the rain drops with respect to the moving girl is: \frac{30}{\sqrt{2}} \, \text{kmh}^{-1}\.

Therefore, the correct answer is \frac{30}{\sqrt{2}} \, \text{kmh}^{-1}\.

The Correct Option is C