Equation of Trajectory MCQ Quiz in বাংলা - Objective Question with Answer for Equation of Trajectory - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

• In projectile motion, the horizontal component of velocity (u cosθ), acceleration (g), and mechanical energy remain a constant while, speed, velocity, the vertical component of velocity (u sinθ), momentum, kinetic energy, and potential energy all changes.
• Velocity and KE are maximum at the point of projection while minimum (but not zero) at the highest point.

CALCULATION: 

• Horizontal range: It is the horizontal distance travelled by a body during the time of flight.

The mathematical expression of the horizontal range is:

\(⇒ R = \frac{{{v^2}\sin 2\theta }}{g}=\frac{2v^2\, sin \, \theta\, \, cos\, \theta}{g}\)     ------- (1)

• Maximum height: It is the maximum height from the point of projection, a projectile can reach

The mathematical expression of the horizontal range is:

\(⇒ H = \frac{{{v^2}{{\sin }^2}\theta }}{{2g}}\)     ------ (2)

ATQ,

⇒ R = H

\(\Rightarrow \frac{2v^2\, sin \, \theta\, \, cos\, \theta}{g}=\frac{v^2\, sin^2\theta}{2g}\)

\(\Rightarrow 2\, cos\,\theta=\frac{sin\, \theta}{2}\)

\(\Rightarrow \theta = tan^{-1} (4)\)

Important Points

• Time is taken to reach maximum height: it is half of the total time of flight.

\(⇒ {{\rm{T}}_{1/2}} = \frac{{{\rm{v\;sin\theta }}}}{{\rm{g}}}\)

• Time of Flight: The time of flight of projectile motion is the time from when the object is projected to the time it reaches the surface.

\(⇒ {\rm{T}} = \frac{{2{\rm{\;v\;sin\theta }}}}{{\rm{g}}}\)

Where T is the total time taken by the projectile, g is the acceleration due to gravity.​

CONCEPT:

• Projectile motion: 

​

• Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • Initial Velocity: The initial velocity can be given as x components and y components.

ux = u cosθ

uy = u sinθ

Where u stands for initial velocity magnitude and θ  refers to projectile angle.

EXPLANATION:

• Trajectory: A projectile is an object that is given an initial velocity, and is acted on by gravity. The path the object follows is called its trajectory.
• Equation of trajectory: The trajectory has horizontal (x) and vertical (y) position components.
• If a projectile is launched with an initial velocity v0, at an angle θ from the horizontal plane, then its vertical position can be found from its horizontal position and is given by:

\({\rm{y}} = {\rm{x\;tanθ }} - \frac{{{\rm{g}}{{\rm{x}}^2}}}{{2{\rm{v}}^2{{\cos }^2}{\rm{θ }}}}\)

y = vertical position (m)

x = horizontal position (m)

v0 = initial velocity (combined components, m/s)

g = acceleration due to gravity (9.80 m/s2)

θ = angle of the initial velocity from the horizontal plane (radians or degrees)

CONCEPT:

• When a body is moving in a plane is then the motion is called a two-dimensional motion.
• Projectile Motion is an example of Two-dimensional motion.
• When a body during its motion makes an acute angle with the horizontal surface then the motion is called Projectile Motion.

• Equation of a projectile motion

\(y = x Tan\theta - \frac{gx^{^{2}}}{2u^{2}cos^{2}\theta}\)

EXPLANATION:

• Let us consider the velocity at point A is u = 2î + 3ĵ and velocity at point B is v

​

• We know that during a projectile motion the horizontal component remains the same only the vertical component change with time.
• The horizontal component at point B is 2î.
• Now by seeing the figure, we can say that the direction vertical component of v is on the negative y-axis and the value is the same as of u but with a negative sign.
• The vertical component of the v is -3ĵ.
• So, the velocity at point B is v = 2î =3ĵ.

Hence option 1 is correct.

Additional Information

• Time of flight

​\(T = \frac{2usin\theta}{g} \)

• Range 

​\(R = \frac{u^{2}sin2\theta}{g} \)

• Height 

​\(H = \frac{u^{2}sin^{2}\theta}{2g} \)

CONCEPT:

• When a body is moving in a plane is then the motion is called a two-dimensional motion.
• Projectile Motion is an example of Two-dimensional motion.
• When a body during its motion makes an acute angle with the horizontal surface then the motion is called Projectile Motion.

• Equation of a projectile motion

​​\(y = xTanθ - \frac{gx^{2}}{2u^{2}cos^{2}θ}\)

CALCULATION:

• Given: 

y = x - 2x2/5

• We know the equation of a projectile is 

\(y = xTanθ - \frac{gx^{2}}{2u^{2}cos^{2}θ}\)

• On comparing the given equation with the standard equation we get 

θ = 45° and \(\frac{g}{2u^{2}cos^{2}θ} = \frac{2}{5}\)

• Now put the value of θ = 45° we get 

\(\frac{g}{2u^{2}cos^{2}45 } = \frac{2}{5}\)

u = 5 m/sec

Hence option 4 is correct

Additional Information

• Time Of Flight

\(T = \frac{2usin\theta}{g}\)

• Range

​\(R = \frac{u^{2}Sin2\theta}{g}\)

• Height 

​\(H = \frac{u^{2}sin^{2}\theta}{2g}\)