Relative Motion: Rain Example
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- 45∘
- 90∘
- 30∘
- 15∘
- 3√3 km/h
- 3√2 km/h
- 3√5 km/h
- 5√3 km/h
- √|→Vr|2+|−→Vm|2
- √|→Vr|2+|−→Vm|2+2|→Vr||−→Vm|cosθ
- √|→Vr|2+|−→Vm|2−2|→Vr||−→Vm|cosθ
- √|→Vr|2−|−→Vm|2
- 12√3 km/h
- 6√3 km/h
- 4√3 km/h
- 2√3 km/h
At a certain place, horizontal component is times the vertical component, the angle of dip at this place is.
none of these
- 5 km/hr
- 4km/hr
- 1km/hr
- 3 km/hr
A man sitting in a bus traveling in a direction from West to East with a speed of km/hr observes that the raindrops are falling vertically down. To another man standing on the ground, the rain will appear
To fall vertically down
To fall at an angle going from West to East
To fall at an angle going from East to West
The information given is insufficient to decide the direction of rain
- 2√7
- 4√7
- 5√7
- 6√7
- 20 kmh−1
- 10√2 kmh−1
- 20√3 kmh−1
- 10 kmh−1
- It is not possible
- Speed of the rain relative to the ground is 2 m/s
- Speed of the man when he finds rain to be falling at angle 45∘ with the vertical, is 4 m/s
- The man has travelled a distance 16 m on the road by the time he again finds rain to be falling at angle 45∘
- 45∘
- 90∘
- 30∘
- 15∘
- 5 km/hr
- 4km/hr
- 1km/hr
- 3 km/hr
- 5 km/h
- 10 km/h
- 10√2 km/h
- 20 km/h
From the previous problem at what angle to the horizontal must be steered in order to reach a point on the opposite bank directly east from the starting point? (Assuming his speed w.r.t. to the river is 4 m/s)
0∘
30∘
45∘
60∘
- 5√5 m/s, θ=tan−142
- 5√5 m/s, θ=tan−112
- 10 m/s, θ=tan−12
- 10 m/s, θ=tan−112
Rain is falling vertically with a speed of 12ms−1.A cyclist is moving east to west with a speed of 12√3ms−1.In order to protect himself from rain at what angle he should hold his umbrella w.r.t vertical.
60∘
0∘
30∘
45∘
- 5 ms−1, 37∘ with respect to vertical
- 5 ms−1, 53∘ with respect to vertical
- 5 ms−1, 45∘ with respect to horizontal
- 5 ms−1, 30∘ with respect to horizontal
A flag is hoisted on the steamer boat.On the basis of the data given below find out the direction of flutter of
the flag
→vsteamer/water=10√3ms−1 towards North
→vwater=5ms−1 towards East
→vwind=5ms−1 towards East
(Assume the wind does not affect the steamer)
North
South
East
West
- 150∘
- 120∘
- 90∘
- 60∘
A man wishes to cross a river in a boat. If he crosses the river in minimum time he takes 10 minutes with a drift of 120 m. If he crosses the river taking shortest route, he takes 12.5 minutes, find velocity of the boat with respect to water.
12 m/min
20 m/min
5 m/min
31 m/min
- √116 m/s
- √32 m/s
- √10 m/s
- 5 m/s
- 3√2 km/h
- 4√2 km/h
- 2√2 km/h
- 5√2 km/h
- 20 m/s, 10 m/s
- 10 m/s, 20√3 m/s
- 10√3 m/s, 20 m/s
- 20 m/s, 10√3 m/s
- At an angle tan−1(3√35) with the inclined plane.
- At an angle tan−1(3√35) with the horizontal.
- At an angle tan−1(√37) with the inclined plane.
- At an angle tan−1(√37) with the vertical.
A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of 2 m/s. At what angle α with the vertical should the screen be placed so that the rain drops falling downwards with velocity 6 m/s strike the windscreen perpendicularly?
tan−1(13)
tan−1(3)
cos−1(3)
sin−1(13)
- 8 km/h
- 7 km/h
- 6 km/h
- 5 km/h
- √|→Vr|2−|−→Vm|2
- √|→Vr|2+|−→Vm|2
- √|→Vr|2+|−→Vm|2+2|→Vr||−→Vm|cosθ
- √|→Vr|2+|−→Vm|2−2|→Vr||−→Vm|cosθ
- 5 km/h
- 10 km/h
- 10√2 km/h
- 20 km/h
- 5 ms−1, 37∘ with respect to vertical
- 5 ms−1, 53∘ with respect to vertical
- 5 ms−1, 45∘ with respect to horizontal
- 5 ms−1, 30∘ with respect to horizontal