Range on an Incline
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- 18 cm
- 14 cm
- 27 cm
- 20 cm
- √3 ut2
- √3 ut
- 2 ut
- ut√3
- 20, 000 N/C
- 10, 000 N/C
- 40, 000 N/C
- 90, 000 N/C
- v2g(1+sinθ)
- v2g
- v2g(1−sinθ)
- v2g(1+cosθ)
[here T1 & T2 are times of flight in the two cases respectively]
- h1=h2
- R2−R1=T21
- R2−R1=g sin θT22
- R2−R1=g sin θT21
- α+β=45∘
- α+β=30∘
- α+β=90∘
- α−β=30∘
- v2g
- v2g(1+sinθ)
- v2g(1+cosθ)
- v2g(1−sinθ)
- R1+R22
- R1R2R1+R2
- 2R1R2R1+R2
- R1+R2R1R2
- 2u2gcosα
- 2u2gtanα
- 2u2gtanαcosα
- 2u2gtanαsinα
- √2[2v2g]
- v2g
- √2v2g
- 2v2g
Find the range of the particle on the plane when it strikes the plane. (Assume the incline is long enough and g=10 m/s2)
- 3.5 m
- 4 m
- 4.2 m
- 5 m
- 30 m
- 60 m
- 90 m
- 150 m
- The direction of projection to get maximum range divides the angle between the vertical and the inclined plane in the ratio of 1:2.
- The direction of projection to get maximum range divides the angle between the vertical and the inclined plane in the ratio of 2:1.
- The direction of projection to get maximum range divides the angle between the vertical and the inclined plane in the ratio of 1:1.
- The direction of projection to get maximum range divides the angle between the vertical and the inclined plane in the ratio of 3:1.
- θ0=cos−11√b2+1, v0=√g2b(1+a2)
- θ0=cos−11√a2−1, v0=√g2b(1+a2)
- θ0=cos−11√a2+1, v0=√g2b(1+a2)
- θ0=cos−11√a2+1, v0=√g2a(1+b2)
- x=80√3[√3−1] m
- x=40√3[√3−1] m
- x=403[√3−1] m
- x=803[√3−1] m
- 26 cm
- 20 cm
- 18 cm
- 14 cm
- R1+R22
- R1R2R1+R2
- 2R1R2R1+R2
- R1+R2R1R2
A ball is projected with a speed at an angle of with the horizontal, from the top of a cliff as shown in the figure. The ball has been thrown with the aim so that it just hits directly the edge of the cliff, but the ball just misses the target and falls on ground as shown in figure. [Take .]
Based on above information, answer the following question:
The value of is
A particle is projected at an angle of 37∘ with an inclined plane. The inclined plane is at an angle of 60∘ with the horizontal. Find
I. Time of flight of particle.
II. Distance traveled by particle (AB) along the inclined plane
I. Time of flight =4.8s
II. Distance traveled =11.4mI. Time of flight =2.4s
II. Distance traveled =5.7mI. Time of flight =1.2s
II. Distance traveled =5.7mI. Time of flight =6.4s
II. Distance traveled =11.4m
If and represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is
- √3uT2
- 2ut
- √3uT
- uT√3
- zero
- tan−1α
- tan−1α+gsinαgcosα
- tan−1ag
[here T1 & T2 are times of flight in the two cases respectively]
- h1=h2
- R2−R1=T21
- R2−R1=g sin θT22
- R2−R1=g sin θT21
- 8ms−1 upward
- −8ms−1 downward
- 32ms−1 upward
- 32ms−1 downward
- x=80√3[√3−1] m
- x=40√3[√3−1] m
- x=403[√3−1] m
- x=803[√3−1] m
- a2b2c2R2
- a2b2c24R2
- 4a2b2c2R2
- a2b2c28R2