Time of Flight with Ground as Frame of Reference
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The position of a body moving along -axis at time is given by . The distance traveled by the body in time interval to .
- 1 s
- 2 s
- 2√3 s
- 4√3 s
A car made a run of in hours. If the speed had been more, it would have taken less for the journey. Find
- √32 s
- 2√3 s
- √3 s
- 0.5 s
- (3, 0)
- (4, 0)
- (2, 0)
- (2√5, 0)
A particle is travelling along a straight line . The distance of the particle from at a time is given by , where is time in . The distance of the particle from when it comes to rest is
- 21→i+14→j+42→k
- 3→i+2→j+6→k
- 3→i+2→j+6→k7
- −( 3→i+2→j+6→k)
A particle is projected at an angle of 37∘ with an inclined plane as shown in figure. Calculate:
(i) Time of flight of particle.
(ii) Distance traveled by particle (AB) along the inclined plane
Time of flight = 2 sec
Distance along incline = 16 m
Time of flight = 3 sec
Distance along incline = 6 m
Time of flight = 125sec
Distance along incline = 125(8−6√3)m
Time of flight = 125sec
Distance along incline = (8−6√3)m
- √Hcosθ/g
- √2Hcosθ/g
- √4H/g
- √8H/g
If a particle is projected from point A, normal to the plane calculate
(i) Time of flight?
(ii) AB=?
Height of point of projection should be given
(Take g=10 m/s2)
- 20 cm
- 180 cm
- 26 cm
- 14 cm
- 9.8m
- 19.6m
- 4.9m
- 39.2m
- 0 ms−1
- 40 ms−1
- 9.8 ms−1
- ∞
- 15o
- 30o
- 60o
- 45o
- 7s
- 5s
- 9s
- 2s
- 3m
- 4m
- 2.4m
- 5m
- 2√20 m/sec, 60
- 20√3 m/sec, 60
- 6√40 m/sec, 30
- 40√6 m/sec, 30
- 2√3 s
- √32 s
- √3 s
- 0.5 s
- 250 m
- 500 m
- 750 m
- 1000 m
If a particle is projected from point A, normal to the plane calculate
(i) Time of flight?
(ii) AB=?
T=2 secAB=40√3 m
T=1 secAB=20√3 m
Height of point of projection should be given
T=4 secAB=40√3 m
If a particle is projected from point A, normal to the plane calculate
(i) Time of flight?
(ii) AB=?
T=2 secAB=40√3 m
T=1 secAB=20√3 m
Height of point of projection should be given
T=4 secAB=40√3 m
- HR=4cotθ
- RH=4cotθ
- HR=4tanθ
- RH=4tanθ
- 3vsinθ
- vsinθ
- vsinθ√2
- vsinθ√3
- (9√3+12)2−√3m
- (30√3+51)m
- (60+30√3)m
- 30(2−√3)m