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A girl standing on a road holds her umbrella at 45 degrees with the vertical to keep rain off. She then runs at a speed of 15*sqrt(2) km/h without the umbrella and finds the rain strikes her head vertically. What is the speed of the raindrops relative to the running girl?

1. 30 km/h
2. 25/sqrt(2) km/h
3. 30/sqrt(2) km/h
4. 25 km/h

Correct answer: 30/sqrt(2) km/h

Solution

When standing with umbrella at 45 deg from vertical, the rain velocity in ground frame makes 45 deg with vertical. So if rain speed components are (vₓ horizontal, v_y vertical), then vₓ/v_y = tan(45 deg) = 1, meaning vₓ = v_y. When the girl runs at 15*sqrt(2) km/h horizontally, rain appears to fall vertically on her. This means the horizontal component of rain velocity relative to her is zero: vₓ - 15*sqrt(2) = 0, so vₓ = 15*sqrt(2) km/h. Therefore v_y = 15*sqrt(2) km/h. Speed of rain relative to running girl = v_y (only vertical since horizontal component cancels) = 15*sqrt(2) km/h vertically... Wait, let me recheck. Velocity of rain relative to girl = (vₓ - 15*sqrt(2)) i + (-v_y) j = 0*i + (-15*sqrt(2)) j. So magnitude = 15*sqrt(2) km/h. But that is not 30 km/h. Let me reconsider. If rain falls at 45 deg to vertical from standstill: tan(45) = v_horizontal/v_vertical. When she runs at u = 15*sqrt(2) km/h and rain appears vertical: v_horizontal = 15*sqrt(2). So v_vertical = 15*sqrt(2). Rain speed relative to running girl: only vertical component remains = 15*sqrt(2) * sqrt(2)... no. Relative velocity = (v_horizontal - u) i - v_vertical j = 0 i - 15*sqrt(2) j. Magnitude = 15*sqrt(2) km/h. Hmm, but that is not one of the options. Let me re-read: rain falls at 45 deg so vₓ = v_y. She runs at 15*sqrt(2) km/h. Rain appears vertical when running, so vₓ = 15*sqrt(2). Then v_y = 15*sqrt(2). Relative velocity to girl = sqrt((vₓ - 15*sqrt(2))² + v_y²) = sqrt(0 + (15*sqrt(2))²) = 15*sqrt(2) km/h. Not 30. But the correct answer should be 30 km/h. Let me reconsider the geometry. If umbrella is at 45 deg with vertical, rain makes 45 deg with vertical (toward her front), so v_rain = v_y downward + vₓ forward where vₓ/v_y = tan 45 = 1. She runs at 15*sqrt(2) km/h forward. Rain appears vertical: this means vₓ = 15*sqrt(2). So v_y = 15*sqrt(2). Speed of rain in ground frame = sqrt(vₓ² + v_y²) = sqrt(2)*15*sqrt(2) = 30 km/h. Speed of rain relative to girl = v_y (vertical only) = 15*sqrt(2) km/h. But option says 30 km/h... The question asks speed of rain relative to running girl. Relative vel = v_rain - v_girl = (15*sqrt(2) i - 15*sqrt(2) j) - (15*sqrt(2) i) = -15*sqrt(2) j. Magnitude = 15*sqrt(2). So the answer is 30/sqrt(2) = 15*sqrt(2) km/h which matches option 30/sqrt(2) km/h.

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