Applications of Horizontal and Vertical Components
Trending Questions
[If T is the time of flight]
An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is
1200 m
0.33 km
3.33 km
33 km
(g=10 m/s2)
- v=20√3 m/s
- θ=60∘
- θ=30∘
- v=10√3 m/s
- 3 m/s
- 4 m/s
- 3√2 m/s
- Data insufficient
- 10 m/sec2
- 5 m/sec2
- 20 m/sec2
- 2.5 m/sec2
- True
- False
A ball whose kinetic energy is E, is thrown at an angle of 45∘ with the horizontal. Its kinetic energy at the highest point of its trajectory will be
E
E√2
E2
Zero
- 2m/s
- 3m/s
- 4 m/s
- Data insufficient
A particle P is projected with velocity u1 at an angle of 30∘ with the horizontal. Another particle Q is thrown vertically upwards with velocity u2 from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is
u1=u2
u1=2u2
u1=u22
u1=4u2
- Kinetic energy will be zero at the highest point of the trajectory
- Vertical component of momentum will be conserved
- Horizontal component of momentum will be conserved
- Gravitational potential energy will be minimum at the highest point of the trajectory
A particle of mass 100 g is fired with a velocity 20 m sec−1 making an angle of 30∘ with the horizontal. When it rises to the highest point of its path then the change in its momentum is
√3 kg m sec−1
12kg m sec−1
√2 kg m sec−1
1 kg m sec−1
A projectile thrown with an initial velocity u has the same range R when the maximum height attained by it is either h1 or h2. Then R, h1 and h2 are related as
R=√h1h2
R=2√h1h2
R=3√h1h2
R=4√h1h2
- At h2 from the ground
- At h4 from the ground
- Depends upon mass and volume of the body
- At 3h4 from the ground
A projectile has a range R and time of flight T. If the range is doubled by increasing the speed of projection, without changing the angle of projection, the time of flight will become
T√2
√2 T
T2
2 T
It is possible to project a particle with a given speed in two possible ways so as to have the same range, R. The product of the times taken to reach this point in the two possible ways is proportional to
1/R
R
R3
1R2
एक कण को क्षैतिज के साथ 30º कोण पर 20 m/s के वेग से प्रक्षेपित किया जाता है। कण की चाल किस समय पर न्यूनतम है? [g = 10 m/s2]
- 1 s
- √3s
- 2 s
- √32s
A particle of mass 100 g is fired with a velocity 20 m sec−1 making an angle of 30∘ with the horizontal. When it rises to the highest point of its path then the change in its momentum is
√3 kg m sec−1
12kg m sec−1
√2 kg m sec−1
1 kg m sec−1
- 60∘
- 45∘
- 30∘
- 15∘
The horizontal distance x and the vertical height y of a projectile at a time t are given by
x=at and y=bt2+ct
where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?
√a2+(2b+c)2
√2a2+(b+c)2
√2a2+(2b+c)2
√a2+(b+2c)2
An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is
1200 m
0.33 km
3.33 km
33 km
Column IColumn IIi. Velocity at half of the maximum height in vertical directiona.v2ii Velocity at the maximum heightb.v√2iii. Change in its velocity when it returnsc.v√2to the groundiv. Average velocity when it reaches d.v2√52the maximum height
- i- a ii-b iii- d iv- c
- i- a ii-b iii- c iv- d
- i- b ii-a iii- d iv- c
- i- a ii-d iii- b iv- c
[If T is the time of flight]
A cricketer hits a ball with a velocity 25 m/sat 60∘above the horizontal. How far above the ground it passes over a fielder 50 mfrom the bat (assume the ball is struck very close to the ground)
8.2 m
9.0 m
11.6 m
12.7 m
A boy throws a ball with a velocity u at an angle θ with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this, he should run with a velocity of :
u cos θ
u sin θ
u tan θ
√u2tanθ
- 2m/s
- 3m/s
- 4 m/s
- Data insufficient
- √(v cos θ)2+(v sin θ)2
- √(v cos θ−v sin θ)2−gt
- √v2+g2t2−(2v sin θ)gt
- √v2+g2t2−(2v cos θ)gt
- Yes, 60∘
- Yes, 30∘
- No
- Yes, 45∘
- 20
- 30
- 40
- 50
At what angle should a ball be projected up an inclined plane with a velocity so that it may hit the incline normally. The angle of the inclined plane with the horizontal is α.
θ=cot−1(12cot α)
θ=tan−1(12cot α)
θ=tan−1(12tan α)
θ=cot−1(12tan α)
A particle is projected from a point O with a velocity u in a direction making an angle α upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is
u sin αg
u cosec αg
u tan αg
u sec αg