The statement “velocity and acceleration always act through the same straight line, either in the same or opposite direction” applies for which dimensional motions?
Solution and Explanation
In one-dimensional motion, an object moves along a straight line, and its position, velocity, and acceleration can be described using a single coordinate or axis. The statement implies that in one-dimensional motion, the velocity and acceleration vectors are parallel or antiparallel to each other, meaning they either point in the same direction or in opposite directions along the straight line of motion.
However, in two or three-dimensional motions, such as motion in a plane or in space, the velocity and acceleration vectors can have components in different directions. In these cases, the velocity and acceleration vectors may not always align or have a simple relationship with each other along a single straight line. The relationship between velocity and acceleration in multi-dimensional motion can be more complex and dependent on the specific motion and forces involved.
Top Questions on projectile motion
- A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2. Time required to reach the ground, if it is dropped from the top of the tower, is :
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- The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ______ m.
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- A body of mass \( M \) thrown horizontally with velocity \( v \) from the top of the tower of height \( H \) touches the ground at a distance of \( 100 \, \text{m} \) from the foot of the tower. A body of mass \( 2M \) thrown at a velocity \( \frac{v}{2} \) from the top of the tower of height \( 4H \) will touch the ground at a distance of \(\dots\) \(\text{m}\).
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- The angle of projection for a projectile to have same horizontal range and maximum height is :
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- A body of mass 'm' is projected with a speed 'u' making an angle of \(45^\circ\) with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as \(\frac{\sqrt{2} \, m u^3}{X g}\). The value of 'X' is ______.
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- Solutions
Concepts Used:
Projectile Motion
Projectile
A projectile is an object set in flight by applying an external force. The projectile may be thrown up vertically or at an angle to the horizontal. It may be dropped from a position of rest. The only force acting on the projectile during its motion along the flight path is the gravitational force and it is in motion due to its own inertia.
Examples of projectile are: a ball hit by a bat, bullet fired from a gun, shell launched from a launcher, bomb dropped from a plane, etc. It must be noted that a rocket or a missile cannot be considered as a projectile as they are propelled by power.
Trajectory path/ projectile motion
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion or trajectory motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
- Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
- Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Acceleration in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.
The types of Projectile Motion Formula are:
- Horizontal Distance – x = Vx0t
- Horizontal Velocity – Vx = Vx0
- Vertical Distance, y – Vy0t – ½ gt2
- Vertical Velocity, Vy – Vy0 – gt