The Equation for the Path of Projectile
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- 16 m
- 8 m
- 3.2 m
- 12.8 m
- sin−1(45)
- sin−1(35)
- sin−1(43)
- sin−1(34)
At the top of the trajectory of a projectile, the directions of its velocity and acceleration are
Parallel to each other
Perpendicular to each other
Antiparallel to each other
Inclined to each other at an angle of 45∘
(Take g=10 ms−2)
- tan−123
- tan−132
- tan−174
- tan−145
- y=x−5x2
- y=2x−54x2
- y=4x−5x2
- y=4x−25x2
- y=2x−5x2
- y=x−5x2
- y=2x−1.25x2
- y=2x−25x2
- 83 m
- 43 m
- 34 m
- 38 m
- 2πR2
- 27πR2
- πR2
- 3πR2
1(u2sin2αg−1)
- 1(u2sin2αag−1)
- 1(usin2αag−1)
- 1(u2sin2α2ag−1)
Column IColumn II(A)Angle of projection(p)20 m(B)Angle of velocity(q)80 mwith horizontalafter 4 s(C)Maximum height(r)45∘(D)Horizontal range(s)tan−1(12)
- A→s, B→s, C→q, D→p
- A→r, B→r, C→q, D→p
- A→r, B→r, C→p, D→q
- A→s, B→r, C→p, D→q
- Minimum
- Zero
- g
- Maximum
A ball is thrown vertically upwards with a velocity of What will be the height reached by the body at the end of (consider )?
The path of a particle moving under the influence of a force fixed in magnitude and direction is
Straight line
Circle
Parabola
Ellipse
- √10 m/s, 30∘
- √10 m/s, 45∘
- 10 m/s, 30∘
- 10 m/s, 45∘
A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 602 with the horizontal. Will it hit a vertical wall 5 m away from the point of projection and perpendicular to the plane of project ionm without hitting the floor ? Will the answer differ if the wall is 22 m away ?
- 1:5
- 5:1
- 1:40
- 40:1
(h<<R, where R, where R is the radius of the earth)
- R2ω22g
- R2ω2g
- R2ω24g
- R2ω28g
- 2√g
- 4√g
- 1√g
- 3√g
- tan θ=2dl
- tan θ=4dl
- tan θ=6dl
- tan θ=8dl
- 2√g
- √2√g
- 3√2√g
- 2√2√g
- θ0=cos−11√b2+1, v0=√g2b(1+a2)
- θ0=cos−11√a2+1, v0=√g2b(1+a2)
- θ0=cos−11√a2+1, v0=√g2a(1+b2)
- θ0=cos−11√a2−1, v0=√g2b(1+a2)
- u2x
- xu
- 2ux
- ux
Column IColumn II(a)Range(p)PQ(b)Maximum height(q)P(c)Time of flight(r)P24Q(d)Tangent of angle of projection(s)√2QgP
- a→s, b→q, c→r, d→q
- a→p, b→r, c→q, d→q
- a→p, b→r, c→s, d→q
- a→p, b→q, c→r, d→s
From the top of a tower of height 10 m, one fire is shot horizontally with a speed of 5*root of 3/s. Another fire is shot upwards at an angle of 60 degrees with the horizontal at some interval of time with the same speed of 5*root of 3 m/s. The shots collide in air at a certain point. The time interval between the 2 shots?
- Radius of curvature at firing point is 50 m.
- As projectile moves, its radius of curvature decreases continuosly.
- As projectile moves its radius of curvature first decreases and then it increases.
- Minimum radius of curvature is 30 m.
- y=2x−5x2
- y=x−5x2
- y=2x−1.25x2
- y=2x−25x2
A girl sees through a circular glass slab(refractive index 1.50 of thickness 20 mm and diameter 60 cm to the bottom of a swimming pol. Refractive index of water is 1.33. The bottom surface of the slab is in contact with the water surface.
The depth of swimming pool is 6m. The area of bottom of swimming pool that can be seen through the slab is approximately.
100 m2
160 m2
190 m2
220 m2
A particle moves along parabolic path y=2x-x2 +2 in such a way that the x component of velocity vector remains constant(5m/s). Find the magnitude of acceleration of the particle.
- u=4√5 ms, θ=45∘
- u=4√5 ms, θ=tan−1(3)
- u=2√5 ms, θ=45∘
- u=8√5 ms, θ=tan−1(3)