Projectiles MCQ Quiz in தமிழ் - Objective Question with Answer for Projectiles - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is R.

Key Points

• Range is same for angles of projection θ and (90-θ)
  • ​t₁= 2usinθ/g
  • t₂= [2u sin(90-θ)]/g = 2u cosθ/g.
  • hence t₁t₂= 4(u² sinθcosθ/g²) = 2/g[u²sinθ/g]
  • =(2/g)R, where R is range.
  • hence,t₁t₂ is directly proportional to R because R is constant.

Concept:

Equation of projectile will be of the form

y = Ax + Bx²

This is the equation of the parabola

Derivation:

Let the projectile velocity be U with an angle \(\begin{array}{l} \theta \\ \end{array}\) from the horizontal

Horizontal component of velocity

\(\begin{array}{l} \mathop u\nolimits_x = u\cos \theta \\ \end{array}\)

vertical component of velocity

\(\begin{array}{l} \mathop u\nolimits_y = u\sin \theta - gt\\ \end{array}\)

\(x = u\cos \theta t\)    (1)

\(y = u\sin \theta t - \frac{1}{2}g\mathop t\nolimits^2 \)     (2)

substituting t from (1) in (2), we get

\(\begin{array}{l} y = x\tan \theta - \frac{{g\mathop x\nolimits^2 }}{{2\mathop u\nolimits^2 \mathop {\cos }\nolimits^2 \theta }}\\ \end{array}\)

This is the equation of the parabola

Explanation:

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.

The maximum height is attained at point A  and using the kinematic equation it can be given as:

\({v^2} = {u^2} + 2as\)

where

s = maximum height H

v = Final velocity in the y-direction which is zero for maximum height = 0

u = Component of initial velocity in the y-direction uy = usinθ

g = acceleration due to gravity the body will be decelerated due to gravity hence we took it as negative

\({v^2} = {u^2} + 2as \Rightarrow 0 = u_y^2 - 2gH\)

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

Important Points

1. For a projectile, the maximum height attain will depend on the Y component of its initial velocity (uy).
2. we have to consider the acceleration due to gravity because as a particle is thrown upward due to gravitational acceleration its velocity will keep on decreasing and at extreme position uy becomes zero and body accelerated downward hence it starts falling again
3. we took acceleration an as acceleration due to gravity g, whereas the body will be decelerated due to gravity hence we took it as negative.
4. Initial Velocity: The initial velocity can be given as x components and y components.
   • Component of initial velocity in x-direction, (ux) = ucosθ
   • Component of initial velocity in the y-direction, (uy) = usinθ
5. Time of flight, \(t = \frac{{2{u}\sin \theta }}{g}\)
6. Horizontal range, \(R = \frac{{u ^2\sin 2\theta }}{g}\)

Concept:

Projectile motion: A type of motion that an object experiences when it is thrown toward the surface of the earth and proceeds along a curved route while being pulled by gravity.

Equation of trajectory:

• \(y = x\tan \alpha -\dfrac{gx^2}{2u^2cos^{2}\alpha }\)

Height of the projectile:

• \(H = \dfrac{{{u^2}{{\sin }^2}\alpha }}{{2g}}\)

Range of the projectile:

• \(R=\dfrac{u^2\sin 2\alpha}{g}\)

Time of flight:

• \(T=\dfrac{2u\sin\alpha}{g}\)

Calculation:

We have, the equation of the trajectory of a projectile given by:

\(y = x\tan \alpha -\dfrac{gx^2}{2u^2cos^{2}\alpha }\)....(i)

Let \(h = \dfrac{u^2 cos^2 \alpha}{2g}\) and \(k = \dfrac{u^2 \sin 2 \alpha}{g}\)

Multiplying (i) by \(\dfrac{2u^2 cos^2 \alpha}{g}\), we get:

⇒ \( \dfrac{2u^2 cos^2 \alpha}{g}× y =\dfrac{2u^2 cos^2 \alpha}{g}× \left(x\tan \alpha -\dfrac{gx^2}{2u^2cos^{2}\alpha }\right)\)

⇒ 4hy = kx - x²

⇒ x² - kx = - 4hy

⇒ \( x^2-2\cdot\frac{k}{2}\cdot x+\frac{k^2}{4} = -4hy+\frac{k^2}{4}\)

⇒ \( \left(x-\frac{k}{2}\right)^2=-4h\left(y-\frac{k^2}{16h} \right)\)

∴ It represents a parabola in the standard form.

So, the length of the latus rectum is equal to 4h and the vertex is \( \left(\dfrac{k}{2}, \dfrac{k^2}{16h} \right)\)

∴ Latus-rectum = 4h

= 4 × \(\dfrac{u^2 cos^2 \alpha}{2g}\)

= \(\dfrac{2u^{2} \cos^{2} \alpha}{g}\)

CONCEPT: 

• Projectile motion: A kind of motion that is experienced by an object when it is projected near the Earth's surface and moves along a curved path under the action of gravitational force.

Given that initial velocity is u. So its vertical component will be u sinθ

The time required to move from O to B is the time of flight T

The time required to move from O to A is equal to from A to B in the figure. Let it is t

T = 2t

From O to A in the vertical direction 

v = u + at

at highest point v = 0

0 = u sinθ - gt

t = u sinθ/g

total time of flight T = 2t​

\(T = \frac{2u~sinθ}{g}\)

The time of flight of projectile is given by:

\(T = \frac{2usinθ}{g}\)

where u is the velocity, makes an angle 'θ' with the x-axis, and g is the gravitational acceleration.

Concept

• Projectile motion: When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and vertical directions. This type of motion is called projectile motion.

\(Total\;time\;of\;flight = \frac{{2\;u\;sinθ }}{g}\)

\(Range\;of\;projectile = \frac{{{u^2}\sin 2θ }}{g}\)

\(Maximum\;Height = \frac{{{u^2}{{\sin }^2}θ }}{{2g}}\)

Where, u = projected speed

θ = angle at which an object is thrown from the ground.

g = acceleration due to gravity = 9.8 m/s2

Maximum Range: It is the longest distance covered by the object during projectile motion.

• When the angle of projection is 45°, the maximum horizontal range is obtained.

Additional Information

• Maximum height is the maximum vertical distance travelled by the projectile from the horizontal plane.

 

We know that,

\(Maximum\;Height = \frac{{{u^2}{{\sin }^2}θ }}{{2g}}\)

For maximizing the vertical range, sin θ must be maximum

Which is sin θ = 1

⇒ θ = 90° 

\(Maximum\; Vertical\,Height = \frac{{u^2}}{{2g}}\)

Hence for vertical maximum range angle should be 90° 

Concept:

• Maximum height: It is the maximum height from the point of projection, a projectile can reach
• The mathematical expression of the height is -
  \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

where u  = initial velocity, θ = angle of projection, H = height reached by the object and g = acceleration due to gravity

The mathematical expression of the height is -

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

The maximum height reached by a body will be when sinθ is maximum i.e. θ = 90°

\(∴ {H_{max}} = \frac{{{u^2}}}{{2g}}\)

Calculation:

Given:

initial velocity (u) = 9.81 m/s,

Maximum height from ground is,

∴ \(∴ {H_{max}} = 25+\frac{{{9.81^2}}}{{2g}}=29.905~m\)

Time of flight: \(t = \frac{{2{v_o}\sin \theta }}{g}\)

Maximum height: \(H = \frac{{v_o^2{{\sin }^2}\theta }}{{2g}}\)

Horizontal Range: \(R = \frac{{v_o^2\sin 2\theta }}{g}\)

When θ = 45°: \({R_{max}} = \frac{{v_o^2}}{g}\)

Explanation:

Projectile motion:

• If a particle is projected in the air with some oblique angle then the particle traces a path and falls on the ground at some point. The particle is called a projectile and its motion in the air is called projectile motion.
• The path traced by a projectile in space is known as trajectory.

The equation of trajectory for the projectile is given by

\({\rm{Y}} = {\rm{X}}\tan {\rm{\alpha }} - \frac{{{\rm{g}}{{\rm{X}}^2}}}{{2{{\rm{u}}^2}{{\cos }^2}{\rm{\alpha }}}}\)

Where,

u = Velocity of projection

α = Angle of projection

The equation is in the form Y = AX + BX2

Where Y = AX + BX2 is the equation of parabola

Hence the path traced by a projectile is parabolic

Important point:

Terms related to projectile motion

1)Time of flight:

It is the duration of time for which a projectile remains in the air

\({\rm{T}} = 2{\rm{u}}\sin \frac{{\rm{\alpha }}}{{\rm{g}}}\)

2) Horizontal range:

It is the horizontal distance between the point of projection and the point of landing.

\({\rm{R}} = \frac{{{{\rm{u}}^2}\sin 2{\rm{\alpha }}}}{{\rm{g}}}\)

3) Height:

It is the maximum vertical distance traveled by the projectile.

\({\rm{H}} = \frac{{{{\rm{u}}^2}{{\sin }^2}{\rm{\alpha }}}}{{2{\rm{g}}}}\)

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.

• For a projectile, the maximum height it can attain will depend on the Y component of its initial velocity (uy).
• Also, we have to consider the acceleration due to gravity, because as a particle is thrown upward due to gravitational acceleration its velocity will keep on decreasing and at extreme position uy becomes zero and body accelerated downward hence it starts falling again
• Now mathematically by using the kinematic equation we can derive the maximum height and it can be given as

\({v^2} = {u^2} + 2as \Rightarrow 0 = u_y^2 - 2gH\)

• Here we took displacement S as maximum height projectile can attain also we took acceleration an as acceleration due to gravity g, whereas the body will be decelerated due to gravity hence we took it as negative
• Initial Velocity: The initial velocity can be given as x components and y components.
  • Component of initial velocity in x-direction, (ux) = ucosθ
  • Component of initial velocity in the y-direction, (uy) = usinθ
• Using this relation, we can derive the relation of maximum height a projectile can attain and it can be given as

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

Time of flight, \(t = \frac{{2{u}\sin \theta }}{g}\)

Horizontal range, \(R = \frac{{u ^2\sin 2\theta }}{g}\)

Concept:

Calculation:

Given that,

A particle having horizontal range (R) is twice the greatest height (H) attained by particle.

This mean that, R = 2H

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

∴ sin 2θ = sin² θ ⇒ 2sin θ.cos θ = sin² θ

tan θ = 2

also \(tan~\theta =\frac{opp~side}{adj~side}=\frac{2}{1}\)

now using Pythagoras theorem as shown below we can find sin θ and cos θ

Therefore, \(\sin \theta =\frac{opp~side}{hyp~side}=\frac{2}{\sqrt{5}}\) similarly \(\cos \theta =\frac{adj~side}{hyp~side}=\frac{1}{\sqrt{5}}\)

Hence range of projectile is given as

\(R=\frac{{{v}^{2}}~sin2\theta }{g}=\frac{{{v}^{2}}\times 2\sin \theta \cos \theta }{g}\)

\(\therefore ~R~=2{{v}^{2}}\left( \frac{2}{\sqrt{5}} \right)\left( \frac{1}{\sqrt{5}} \right)=\frac{4{{v}^{2}}}{5g}\)