Time of Flight with Incline as Frame of Reference
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( Take g=10 m/s2)
- 60∘
- 30∘
- 75∘
- 45∘
- 60∘
- 30∘+tan−1(√32)
- 30∘+tan−1(12√3)
- 45∘
- T=2usinθg
- T=2ucosθg
- T=2utanθg
- T=usinθg
- 2.1 m/s2
- 2.5 m/s2
- 2.7 m/s2
- 2.9 m/s2
(Take g=10 m/s2)
- √3 s
- 1√3 s
- 2√3 s
- 2 s
A copper rod of length L and mass m is sliding down a smooth inclined plane of inclination θ with a constant
speed v. A current I is flowing in perpendicular to the plane of diagram inwards. A vertically upward magnetic
field →B exist in this region.The magnitude of the required magnetic field →B is
mgILsinθ
mgILsinθ
mgILtanθ
mgILsinθ
A bullet fired at an angle of with the horizontal hits the ground away. By adjusting its angle of projection, can one hope to hit a target away? Assume the muzzle speed to be fixed, and neglect air resistance.
- 15 ms−1
- 20 ms−1
- 25 ms−1
- 30 ms−1
A ball A is projected from O with an initial velocity in a direction above the horizontal. Another ball B, from O on a line above the horizontal is released from rest at the instant A starts, as shown in figure.
[Take ]
Based on above information, answer the following questions:
How far will B have fallen when it is hit by A?
- √e2a2+b2 at angle tan−1(aeb) to the vertical
- √a2+e2b2 at angle tan−1(aeb) to the vertical
- √a2+b2e2 at angle tan−1(eab) to the vertical
- √a2e2+b2 at angle tan−1(aeb) to the vertical
- 3m
- 4m
- 5m
- zero
- 1(u2sin2αag−1)
- 1(u2sin2α2ag−1)
- 1(usin2αag−1)
1(u2sin2αg−1)
- 30∘+tan−1(12√3)
- 45∘
- 60∘
- 30∘+tan−1(√32)
- 40 km / hr
- 10 km / hr
- 20 km / hr
- 30 km / hr
(Take g=10 m/s2)
- √3 s
- 1√3 s
- 2√3 s
- 2 s
- 12.6m
- 14.3m
- 17.3m
- 15.4m
- 60∘
- 0∘
- 90∘
- 30∘
- √2sec
- 2√2sec
- 2sec
- none
- 30∘+tan−1(12√3)
- 45∘
- 60∘
- 30∘+tan−1(√32)
- √30m/s
- √15m/s
- 0
- −√15m/s
- RA=RB=RC
- RA=RC>RB
- RA<RB<RC
- RA=RC<RB
- √32 ms−1
- √116 ms−1
- 5 ms−1
- √73 ms−1
- 4.9ms−2 in horizontal direction
- 4.9ms−2 in vertical direction
- 9.8ms−2 in vertical direction
- zero
A particle is projected from the ground with an initial speed of 'v' at angle θ with horizontal. the average velocity of the particle between its point of projection and height point of trajectory is
- v2√1+2cos2θ
- v2√1+cos2θ
- vcosθ
- v2√1+3cos2θ
- y=tanx
- y=αcosωxν0
- y=αsinωxν0
- y=cotx
- (θ+β)<90o, if the wall is smooth
- if the wall is rough, coefficient of restitution <tanβcosθ
- if the wall is rough, coefficient of restitution =tanβcosθ
- none of the above
- gT2/2R
- gT/2R
- T2/Rg
- T2/2Rg