Applications of Horizontal and Vertical Components
Trending Questions
- 1.25 m
- 2.50 m
- 3.75 m
- 4.00 m
A projectile is fired with velocity u making angle θ with the horizontal. What is the change in velocity when it is at the highest point
u cos θ
u
u sin θ
(u cos θ−u)
- tanθ=13
- tanθ=12
- tanθ=2√3
- tanθ=12√3
A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the center of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of Projection and the edge of the boat are in the same horizontal level.
,
,
,
,
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)
12.7 m
8.2 m
9.0 m
11.6 m
(g=10 m/s2)
- v=20√3 m/s
- θ=60∘
- θ=30∘
- v=10√3 m/s
A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30∘ with the horizontal. The change in momentum (in magnitude) of the body is
24.5 N–s
49.0 N–s
98.0 N–s
50.0 N–s
A projectile can have the same range for two angles of projection. if & be the times of flight in the two cases then what is the product of the two times of flight?
The speed of a projectile at the highest point becomes 1√2times its initial speed. The horizontal range of the projectile will be
- 20√3 m/s, 60∘
- 6√40 m/s, 30∘
- 40√6 m/s, 30∘
- 2√20 m/s, 60∘
- Yes, 60∘
- Yes, 30∘
- No
- Yes, 45∘
- 7.5 m/sec at 30∘ clockwise from horizontal
- 15 m/sec at 30∘ clockwise from horizontal
- 7.5 m/sec at 60∘ clockwise from horizontal
- 15 m/sec at 60∘ clockwise from horizontal
The aeroplane is shown here on level flight at an altitude of and at a speed of .
At what distance should it release a bomb to hit the target? [Take ]
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
(Take g=10 ms−2)
- 5 m from A
- 5 m from B
- 2 m from A
- 2 m from B
A body is projected up a smooth inclined plane (length = 20√2m ) with velocity u from the point M as shown in the figure. The angle of inclination is 45∘ and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v
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
- 3 m/s
- 4 m/s
- 3√2 m/s
- Data insufficient
The speed of projection of a projectile is increased by 5%, without changing the angle of projection. The percentage increase in the range will be
7.5%
10%
5%
2.5%
- √(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
A projectile thrown at an angle of 30∘ with the horizontal has a range R1. Another projectile thrown, with the same velocity, at an angle 30∘ with the vertical, has a range R2. The relation between R1 and R2 is
R1=R22
R1=R2
R1=2R2
R1=4R2
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
R
1/R
R3
1R2
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
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
- 10 m/sec2
- 5 m/sec2
- 20 m/sec2
- 2.5 m/sec2