Question 1

Rain is falling vertically at 12 m/s. A man walks along a straight road at 9 m/s. At what angle from the vertical must he hold his umbrella to keep the rain off?

Accepted Answer

tan⁻¹(3/4) ≈ 37° from the vertical. The rain appears tilted by tanθ = v(man)/v(rain) = 9/12 = 3/4, so θ = tan⁻¹(0.75) ≈ 37° from the vertical, tilted forward.

Question 2

Rain falls vertically downward at 20 m/s. A cyclist rides on a level road at 15 m/s. What is the speed of the rain relative to the cyclist?

Accepted Answer

25 m/s. Relative speed = √(v(rain)² + v(cyclist)²) = √(20² + 15²) = √625 = 25 m/s.

Question 3

Rain is coming straight down at 24 km/h. A woman runs at 10 km/h. What is the apparent speed of the rain with respect to her?

Accepted Answer

26 km/h. Apparent speed = √(24² + 10²) = √(576 + 100) = √676 = 26 km/h.

Question 4

A cyclist moving at 15 km/h finds that rain, which is actually falling vertically, appears to strike him at 37° to the vertical. What is the true speed of the rain? (tan 37° = 3/4)

Accepted Answer

20 km/h. tan 37° = v(cyclist)/v(rain), so v(rain) = 15/(3/4) = 20 km/h.

Question 5

To a cyclist, rain appears to fall at 28 m/s making an angle of 30° with the vertical. The rain is actually falling vertically. What is the cyclist’s speed?

Accepted Answer

14 m/s. The horizontal component of the apparent velocity equals the cyclist’s speed: v = 28 sin 30° = 28 × 0.5 = 14 m/s.

Question 6

Rain falls vertically at 21 m/s. A passenger sits in a bus moving at 28 m/s on a straight road. What is the speed of the rain as seen through the bus window?

Accepted Answer

35 m/s. Apparent speed = √(21² + 28²) = √(441 + 784) = √1225 = 35 m/s.

Question 7

To a jogger, rain appears to come at 34 km/h. The rain is actually falling vertically at 30 km/h. How fast is the jogger running?

Accepted Answer

16 km/h. v(jogger) = √(v(apparent)² − v(rain)²) = √(34² − 30²) = √(1156 − 900) = √256 = 16 km/h.

Question 8

A scooter rider moving at 24 km/h finds that vertically falling rain strikes him at exactly 45° to the vertical. What is the speed of the rain?

Accepted Answer

24 km/h. At 45°, tan 45° = v(rider)/v(rain) = 1, so v(rain) = v(rider) = 24 km/h.

Question 9

Rain appears to fall vertically to a man walking at 12 km/h. When he speeds up to 21 km/h in the same direction, the rain appears to meet him at 45° to the vertical. What is the actual speed of the rain?

Accepted Answer

15 km/h. Appearing vertical at 12 km/h means the rain’s horizontal component is 12 km/h; at 21 km/h the relative horizontal speed is 21 − 12 = 9 km/h, and 45° gives vertical component = 9 km/h. True speed = √(12² + 9²) = 15 km/h.

Question 10

Rain is falling at 16 m/s making an angle of 30° with the vertical. At what speed must a person run (in the direction of the rain’s horizontal drift) so that the rain appears to fall vertically on him?

Accepted Answer

8 m/s. He must cancel the horizontal component of the rain: v = 16 sin 30° = 16 × 0.5 = 8 m/s.

Question 11

Rain falls with speed 20 m/s at 37° from the vertical. A boy runs horizontally at just the right speed so that the rain appears vertical to him. What is the apparent speed of the rain for the boy? (sin 37° = 0.6, cos 37° = 0.8)

Accepted Answer

16 m/s. Running at 20 sin 37° = 12 m/s cancels the horizontal part, leaving only the vertical component: 20 cos 37° = 16 m/s.

Question 12

A man walking at 16 km/h finds that vertically falling rain appears to come at 53° to the vertical. What is the apparent speed of the rain with respect to him? (tan 53° = 4/3)

Accepted Answer

20 km/h. Vertical rain speed = 16/tan 53° = 16 × 3/4 = 12 km/h, so apparent speed = √(16² + 12²) = √400 = 20 km/h.

Question 13

Raindrop streaks on the window of a bus moving at 15 m/s make an angle of 37° with the vertical. To a person standing on the road the rain falls vertically. What is the speed of the rain? (tan 37° = 3/4)

Accepted Answer

20 m/s. tan 37° = v(bus)/v(rain) ⇒ v(rain) = 15/(3/4) = 20 m/s.

Question 14

Rain falls vertically at 21 m/s. A car travels at 28 m/s on a straight road. At what angle from the vertical do the raindrops strike the windscreen (as seen by the driver)?

Accepted Answer

tan⁻¹(4/3) ≈ 53°. tanθ = v(car)/v(rain) = 28/21 = 4/3, so θ = tan⁻¹(4/3) ≈ 53° from the vertical.

Question 15

Rain is falling vertically at 12√3 km/h (≈ 20.8 km/h). A runner moves at 12 km/h. What is the speed of the rain relative to the runner?

Accepted Answer

24 km/h. Relative speed = √((12√3)² + 12²) = √(432 + 144) = √576 = 24 km/h.

Question 16

Rain falls vertically at 15 m/s. A motorcyclist finds that the rain strikes him at 53° to the vertical. How fast is the motorcycle moving? (tan 53° = 4/3)

Accepted Answer

20 m/s. tan 53° = v(bike)/v(rain) ⇒ v(bike) = 15 × 4/3 = 20 m/s.

Question 17

Rain pours straight down at 16 m/s while a delivery rider travels at 16 m/s on a level road. What is the magnitude of the rain’s velocity relative to the rider?

Accepted Answer

16√2 ≈ 22.6 m/s. The two perpendicular components are equal, so the relative speed = √(16² + 16²) = 16√2 ≈ 22.6 m/s (at 45° to the vertical).

Question 18

A woman rides a moped at 20 km/h and observes that vertically falling rain approaches her at 53° to the vertical. What is the apparent speed of the rain relative to her? (tan 53° = 4/3)

Accepted Answer

25 km/h. Vertical rain speed = 20/tan 53° = 20 × 3/4 = 15 km/h; apparent speed = √(20² + 15²) = √625 = 25 km/h.

Question 19

To a cyclist riding at 12 km/h, vertically falling rain appears inclined at 30° to the vertical. What is the actual speed of the rain?

Accepted Answer

12√3 ≈ 20.8 km/h. tan 30° = v(cyclist)/v(rain) ⇒ v(rain) = 12/tan 30° = 12√3 ≈ 20.8 km/h.

Question 20

A man walks at 9 km/h and finds that vertically falling rain meets him at 45° to the vertical. What is the speed of the rain relative to him?

Accepted Answer

9√2 ≈ 12.7 km/h. At 45° the vertical rain speed equals his walking speed (9 km/h), so the relative speed = √(9² + 9²) = 9√2 ≈ 12.7 km/h.

Rain Man Problem — IISER Physics MCQs with Solutions

Questions with solutions