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

Last updated on Mar 22, 2025

பெறு Projectiles பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Projectiles MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Projectiles MCQ Objective Questions

Top Projectiles MCQ Objective Questions

Projectiles Question 1:

A projectile can have the same range R for two angles of projection. If t1 and t2 be the time of flights in the two cases, then the product of the two time of flights is proportional to

  1. 1R
  2. R
  3. R2
  4. 1R2

Answer (Detailed Solution Below)

Option 2 : R

Projectiles Question 1 Detailed Solution

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.

Projectiles Question 2:

The shape of the projectile is

  1. a straight line
  2. curvilinear motion
  3. a parabola
  4. a hyperbola

Answer (Detailed Solution Below)

Option 3 : a parabola

Projectiles Question 2 Detailed Solution

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 θ from the horizontal

Horizontal component of velocity

ux=ucosθ

vertical component of velocity

uy=usinθgt

x=ucosθt    (1)

y=usinθt12gt2     (2)

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

y=xtanθgx22u2cos2θ

This is the equation of the parabola

Projectiles Question 3:

A projectile is projected from a point on ground with velocity of projection 'u' and angle of projection 'θ' . How much maximum height can the projectile reach ?

  1. h=usinθ2g
  2. h=u2sin2θ2g
  3. h=u2sinθ2g
  4. h=usin2θ2g

Answer (Detailed Solution Below)

Option 2 : h=u2sin2θ2g

Projectiles Question 3 Detailed Solution

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:

v2=u2+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

v2=u2+2as0=uy22gH

H=uy22g=u2sin2θ2g

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 ubecomes 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=2usinθg
  6. Horizontal range, R=u2sin2θg

Projectiles Question 4:

For a projectile, the true statement is :

  1. the equation of path is y = x tan α − g2x2u2sin2α
  2. the latus rectum of path is 2u2cos2αg
  3. time upto highest point is 2usinαg
  4. maximum range is u2sinαg

Answer (Detailed Solution Below)

Option 2 : the latus rectum of path is 2u2cos2αg

Projectiles Question 4 Detailed Solution

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=xtanαgx22u2cos2α

Height of the projectile:

  • H=u2sin2α2g

Range of the projectile:

  • R=u2sin2αg

Time of flight:

  • T=2usinαg

Calculation:

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

y=xtanαgx22u2cos2α....(i)

Let h=u2cos2α2g and k=u2sin2αg

Multiplying (i) by 2u2cos2αg, we get:

2u2cos2αg×y=2u2cos2αg×(xtanαgx22u2cos2α)

⇒ 4hy = kx - x2

⇒ x2 - kx = - 4hy

x22k2x+k24=4hy+k24

(xk2)2=4h(yk216h)

∴ It represents a parabola in the standard form.

So, the length of the latus rectum is equal to 4h and the vertex is (k2,k216h)

∴ Latus-rectum = 4h

= 4 × u2cos2α2g

= 2u2cos2αg

Projectiles Question 5:

When a projectile is thrown with velocity u, making an angle θ with the horizontal, the total time of flight is

  1. 2u sinθg
  2. 2u sinθ2
  3. 2u sinθ 
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 2u sinθg

Projectiles Question 5 Detailed Solution

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.

F2 J.K Madhu 04.05.20  D3

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=2u sinθg

The time of flight of projectile is given by:

T=2usinθg

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

Projectiles Question 6:

The range of a projectile is maximum, when the angle of projection is -

  1. 60°
  2. 45°
  3. 90°
  4. 30°

Answer (Detailed Solution Below)

Option 2 : 45°

Projectiles Question 6 Detailed Solution

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.

Totaltimeofflight=2usinθg

Rangeofprojectile=u2sin2θg

MaximumHeight=u2sin2θ2g

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,

MaximumHeight=u2sin2θ2g

For maximizing the vertical range, sin θ must be maximum

Which is sin θ = 1

⇒ θ = 90° 

MaximumVerticalHeight=u22g

Hence for vertical maximum range angle should be 90° 

Projectiles Question 7:

A small ball is shot vertically upward from the top of a building 25 m above ground with initial velocity of 9.81 m/sec. Determine the maximum height the ball will reach.

  1. 29.9 m
  2. 40 m
  3. 38.9 m
  4. 7 m

Answer (Detailed Solution Below)

Option 1 : 29.9 m

Projectiles Question 7 Detailed Solution

Concept:

F1 Krupalu Madhu 25.08.20 D4

  • Maximum height: It is the maximum height from the point of projection, a projectile can reach
  • The mathematical expression of the height is -
    H=u2sin2θ2g

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=u2sin2θ2g

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

Hmax=u22g

Calculation:

Given:

initial velocity (u) = 9.81 m/s,

Maximum height from ground is,

∴ Hmax=25+9.8122g=29.905 m

26 June 1

RRB JE ME 45 13Q Full Test 1 Part 5-hindi - Final Diag(Deepak&Shashi) images Q9

Time of flight: t=2vosinθg

Maximum height: H=vo2sin2θ2g

Horizontal Range: R=vo2sin2θg

When θ = 45°: Rmax=vo2g

Projectiles Question 8:

The path traced by a projectile in space is knows as _______.

  1. coral
  2. orbit
  3. track
  4. trajectory

Answer (Detailed Solution Below)

Option 4 : trajectory

Projectiles Question 8 Detailed Solution

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.

F1 Satya Madhu 18.06.20 D9

The equation of trajectory for the projectile is given by

Y=XtanαgX22u2cos2α

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

T=2usinαg

2) Horizontal range:

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

R=u2sin2αg

3) Height:

It is the maximum vertical distance traveled by the projectile.

H=u2sin2α2g

Projectiles Question 9:

A projectile is projected from a point on ground with velocity of projection 'u' and angle of projection 'θ' . How much maximum height can the projectile reach ?

  1. h=usinθ2g
  2. h=u2sin2θ2g
  3. h=u2sinθ2g
  4. h=usin2θ2g
  5. none of the above

Answer (Detailed Solution Below)

Option 2 : h=u2sin2θ2g

Projectiles Question 9 Detailed Solution

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 ubecomes 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


v2=u2+2as0=uy22gH

  • 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=uy22g=u2sin2θ2g

26 June 1

Time of flight, t=2usinθg

Horizontal range, R=u2sin2θg

Projectiles Question 10:

To throw a stone a maximum distance in a projectile motion, the stone needs to be thrown at ________ angle.

  1. 45°
  2. 60°
  3. 10°
  4. 30°

Answer (Detailed Solution Below)

Option 1 : 45°

Projectiles Question 10 Detailed Solution

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.

Totaltimeofflight=2usinθg

Rangeofprojectile=u2sin2θg

MaximumHeight=u2sin2θ2g

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 range is obtained.

We know that,

Range=u2sin2θgAlso,Rmaximum=u2g

Range is maximum when sin 2θ = 1

 θ = 45°

Get Free Access Now
Hot Links: teen patti flush teen patti club apk lotus teen patti