Question 1

Rain is falling vertically with a speed of 15 m/s. A man runs horizontally at 8 m/s on a straight road. What is the speed of the rain relative to the man?

Accepted Answer

17 m/s. Relative to the man the rain has a vertical component 15 m/s and a horizontal component 8 m/s, so v = √(15² + 8²) = √289 = 17 m/s.

Question 2

Rain falls vertically at 24 m/s. To a cyclist riding at 7 m/s on a level road, at what angle with the vertical does the rain appear to fall?

Accepted Answer

tan⁻¹(7/24). In the cyclist frame the rain gains a horizontal component 7 m/s while keeping its vertical component 24 m/s, so tan θ = 7/24 ⇒ θ = tan⁻¹(7/24) from the vertical.

Question 3

A car is moving at 24 m/s on a straight road while rain falls vertically at 7 m/s. With what speed do the raindrops strike the windscreen (speed of rain relative to the car)?

Accepted Answer

25 m/s. Speed of rain relative to the car = √(24² + 7²) = √(576 + 49) = √625 = 25 m/s.

Question 4

To a driver moving at 15 m/s, vertically falling rain appears to have a speed of 17 m/s. What is the actual speed of the rain?

Accepted Answer

8 m/s. Apparent speed² = actual speed² + car speed², so actual speed = √(17² − 15²) = √(289 − 225) = √64 = 8 m/s.

Question 5

Rain falls vertically at 15 km/h. To a person moving on a level road the rain appears to make an angle tan⁻¹(8/15) with the vertical. Find the speed of the person.

Accepted Answer

8 km/h. tan θ = v_person/v_rain, so v_person = 15 × (8/15) = 8 km/h.

Question 6

Rain is falling vertically. To a motorist driving at 20 m/s the drops appear to come at 45° with the vertical. The actual speed of the rain is:

Accepted Answer

20 m/s. tan 45° = v_car/v_rain = 1, so v_rain = 20 m/s.

Question 7

When a motorcyclist moves at 8 m/s along a straight road, rain appears to fall vertically. When he speeds up to 23 m/s in the same direction, the rain appears to make 45° with the vertical. The actual speed of the rain is:

Accepted Answer

17 m/s. Vertical appearance at 8 m/s means the horizontal rain component is 8 m/s; at 23 m/s, tan 45° = (23 − 8)/v_y gives v_y = 15 m/s, so v = √(8² + 15²) = 17 m/s.

Question 8

Rain appears to fall vertically to a cyclist riding at 7 m/s. To a motorist moving at 25 m/s in the same direction on the same road, it appears to make 37° with the vertical (tan 37° = 0.75). Find the vertical component of the actual rain velocity.

Accepted Answer

24 m/s. The horizontal rain component is 7 m/s; tan 37° = (25 − 7)/v_y = 18/v_y = 0.75 ⇒ v_y = 24 m/s.

Question 9

To a girl cycling at 12 km/h, rain appears to fall vertically. When she doubles her speed to 24 km/h, the rain appears to make 45° with the vertical. The actual speed of the rain is:

Accepted Answer

12√2 ≈ 17 km/h. The horizontal rain component is 12 km/h; tan 45° = (24 − 12)/v_y gives v_y = 12 km/h, so v = √(12² + 12²) = 12√2 ≈ 17 km/h.

Question 10

To a passenger in a train running at 20 m/s, rain appears to fall vertically. When the train speeds up to 41 m/s, the rain appears inclined at 45° to the vertical. The true speed of the rain is:

Accepted Answer

29 m/s. The horizontal rain component is 20 m/s; at 41 m/s, tan 45° = (41 − 20)/v_y gives v_y = 21 m/s, so v = √(20² + 21²) = √841 = 29 m/s.

Question 11

Rain is falling at 40 km/h making 37° with the vertical, its horizontal drift pointing east (sin 37° = 0.6). A man runs west at 8 km/h. At what angle with the vertical does the rain appear to fall for him?

Accepted Answer

45°. Rain components: horizontal 40 sin 37° = 24 km/h, vertical 40 cos 37° = 32 km/h; relative horizontal = 24 + 8 = 32 km/h, so tan θ = 32/32 = 1 ⇒ θ = 45°.

Question 12

Rain falls at 20 m/s making 53° with the vertical, drifting in the same direction in which a motorcyclist travels (sin 53° = 0.8). At what speed must he ride so that the rain appears to strike him at 45° with the vertical from the front?

Accepted Answer

28 m/s. Rain components: horizontal 20 sin 53° = 16 m/s, vertical 20 cos 53° = 12 m/s; for a 45° appearance from the front, u − 16 = 12 ⇒ u = 28 m/s.

Question 13

Rain falls at 20 km/h making 37° with the vertical, drifting in the same direction in which a man runs at 24 km/h (the rain comes from behind him, sin 37° = 0.6). At what angle with the vertical should he hold his umbrella?

Accepted Answer

37°, tilted forward (towards his motion). Rain components: horizontal 20 sin 37° = 12 km/h, vertical 16 km/h; relative horizontal = 12 − 24 = −12 km/h (i.e. 12 km/h from the front), so tan θ = 12/16 = 0.75 ⇒ θ = 37° tilted forward.

Question 14

A motorcyclist rides at 12 m/s and vertically falling rain then appears to him to make 37° with the vertical (sin 37° = 0.6). The apparent speed of the rain relative to him is:

Accepted Answer

20 m/s. The horizontal relative component equals his speed: v_app sin 37° = 12 ⇒ v_app = 12/0.6 = 20 m/s.

Question 15

Vertically falling rain appears to a cyclist riding at 24 km/h to make an angle of 53° with the vertical (tan 53° = 4/3). The actual speed of the rain is:

Accepted Answer

18 km/h. tan 53° = v_cyclist/v_rain ⇒ v_rain = 24 ÷ (4/3) = 18 km/h.

Question 16

Rain falling vertically appears to the driver of a car to hit the windscreen at 20 m/s, directed at 53° with the vertical (sin 53° = 0.8). The speed of the car is:

Accepted Answer

16 m/s. The horizontal part of the apparent velocity comes entirely from the car: v_car = 20 sin 53° = 16 m/s (the true rain speed is 20 cos 53° = 12 m/s).

Question 17

The velocity of rain with respect to the ground is (12î − 35ĵ) m/s, where î is horizontal and ĵ points vertically upward. A car moves with velocity 12î m/s. The speed of the rain relative to the car is:

Accepted Answer

35 m/s. v_rel = (12 − 12)î − 35ĵ = −35ĵ, so the rain appears to fall vertically at 35 m/s.

Question 18

Rain is falling vertically at 12 m/s. At what speed must a car travel so that the rain appears to the driver to make 60° with the vertical?

Accepted Answer

12√3 ≈ 21 m/s. tan 60° = v_car/v_rain ⇒ v_car = 12 × √3 = 12√3 ≈ 21 m/s.

Question 19

To a rider, vertically falling rain appears to come at 40 km/h making 53° with the vertical (sin 53° = 0.8). The speed of the rider is:

Accepted Answer

32 km/h. Since the true rain is vertical, the entire horizontal apparent component equals the rider speed: v = 40 sin 53° = 32 km/h.

Question 20

Rain falls vertically at 21 m/s. At what speed should a car move so that to the driver the rain appears inclined at 30° to the vertical?

Accepted Answer

7√3 ≈ 12 m/s. tan 30° = v_car/21 ⇒ v_car = 21/√3 = 7√3 ≈ 12 m/s.

Rain Man Problem — JEE Main Physics MCQs with Solutions

Questions with solutions