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Test: Projectile Motion - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mechanics - Test: Projectile Motion

Test: Projectile Motion for Civil Engineering (CE) 2024 is part of Engineering Mechanics preparation. The Test: Projectile Motion questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Projectile Motion MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Projectile Motion below.
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Test: Projectile Motion - Question 1

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

Detailed Solution for Test: Projectile Motion - Question 1

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

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

  • Time of flight:
    It is the duration of time for which a projectile remains in the air
  • Horizontal range:
    It is the horizontal distance between the point of projection and the point of landing.
  • Height:
    It is the maximum vertical distance traveled by the projectile.
Test: Projectile Motion - Question 2

For a projectile, the true statement is :

Detailed Solution for Test: Projectile Motion - Question 2

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:

Height of the projectile:

Range of the projectile:

Time of flight:

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


∴ It represents a parabola in the standard form.
So, the length of the latus rectum is equal to 4h and the vertex is 
∴ Latus-rectum = 4h

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Test: Projectile Motion - Question 3

When air resistance is neglected, the only force acting on the projectile is ______, which causes the projectile to have a constant downward acceleration.

Detailed Solution for Test: Projectile Motion - Question 3

Projectile motion:

  • In the absence of air resistance there are no forces or components of forces that act horizontally. A velocity vector can only change if there is acceleration (acceleration is the rate of change of velocity). In order to accelerate a resultant force is required (according to Newton's Second Law,  F= ma ).
  • In the absence of air resistance the only force acting on a projectile in flight is the weight of the object. Weight by definition acts vertically downwards, hence no horizontal component. .Therefore , the horizontal motion takes place with constant velocity.
Test: Projectile Motion - Question 4

A bullet is fired upwards at an angle of 30° to the horizontal from a point P on a hill, and it strikes a target which is 80 m lower than P. The initial velocity of the bullet is 100 m/s. Calculate the maximum height to which the bullet will rise above the horizontal. (Assume g = 9.81 m/s2

Detailed Solution for Test: Projectile Motion - Question 4

Projectile Motion:

  • A projectile is any object that once thrown continues in motion by its own inertia.
  • A projectile motion takes place under the influence of the force of gravity only.

Maximum of Projectile height:

  • The maximum height of a projectile is the vertical distance at which the projectile reaches zero vertical velocity.
  • It is given by the equation:


Range of projectile:

  • The range of a projectile is the horizontal distance covered by the projectile during its motion. It is given by the equation:

Where u is the velocity with which the projectile is thrown, θ is the angle at which the projectile is thrown.
Calculation:
Given:
θ = 30°, u = 100 m/s

Test: Projectile Motion - Question 5

To take the longest possible jump, an athlete should make an angle of _____

Detailed Solution for Test: Projectile Motion - Question 5

Projectile motion: 
When a particle is projected obliquely near the earth's surface, it moves simultaneously in horizontal and vertical directions. The path of such a particle is called projectile and the motion is called projectile motion.

Range of projectile:

  • The horizontal range of a projectile is the distance along the horizontal plane it would travel, before reaching the same vertical position as it started from.

Formulae in projectile motion:

where u = projected speed, θ = angle at which an object is thrown from the ground and g = acceleration due to gravity = 9.8 m/s2.
Calculation:
Given:
Range of a Projectile motion is given by (R):

For horizontal distance to be maximum:
sin 2θ = 1
∴ sin 2θ = sin 90° 
∴ θ = 45°.
∴ to take the longest possible jump, an athlete should make an angle of 45° with the ground.

Test: Projectile Motion - Question 6

A projectile is fired at an angle of 30° from horizontal with a speed of Vo m/s. The maximum height attained by the projectile is

Detailed Solution for Test: Projectile Motion - Question 6

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.

  • Maximum Height: The maximum height is reached when vy = 0.

Where h is the maximum height.

CALCULATION:
Given that u = Vo m/s, θ = 30°
Maximum Height 

Test: Projectile Motion - Question 7

A particle is projected at an angle θ to the horizontal and it attains a maximum height H. The time taken by the projectile to reach the highest point, of its path is

Detailed Solution for Test: Projectile Motion - Question 7

At the highest point of the projectile:
For projectile motion we know: ax = 0 and ay = -g
Diagram

Explanation:
Vy = 0 (at highest point)

⇒ Time taken by the projectile  to reach the highest point,

Test: Projectile Motion - Question 8

The direction of projection for the range of a projectile to be maximum on an inclined plane having angle of inclination 30° to the horizontal should be

Detailed Solution for Test: Projectile Motion - Question 8

For the maximum range in case of the inclined plane, the angle from the inclined should be,

where, θ = angle of projection from the inclined plane and α is the angle of incline from horizontal.
Calculation:
Given:
The angle of incline, α = 30°
For maximum range,

θ = 30° with the inclined plane.
So, the angle of projection from the vertical =  90° - ( 30° + 30°) =  30°

Test: Projectile Motion - Question 9

A particle is projected with velocity u at an inclination θ with the horizontal. Then Maximum height (H) attained is

Detailed Solution for Test: Projectile Motion - Question 9

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.

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

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

Where T1/2 = time taken by the projectile to reach maximum height, g = acceleration due to gravity, and v = velocity
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.

Where T is the total time taken by the projectile, g is the acceleration due to gravity.
Range: The range of the motion is fixed by the condition y = 0.

Where R is the total distance covered by the projectile.

Test: Projectile Motion - Question 10

A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)

Detailed Solution for Test: Projectile Motion - Question 10

Projectile motion:

  • It 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.
  • Component of initial velocity in x-direction, (ux) = u cos θ
  • Component of initial velocity in the y-direction, (uy) = u sin θ
  • In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
  • If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.

  • And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:

The height of projectile is given by,

  • Whereas the time of flight is the total time for which projectile stayed in the air.

Time of flight for the projectile,

where The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R
Calculation:
Given:
Height of the tower H1 = 60 m, initial velocity of the object u = 40 m/s, angle of inclination with horizontal θ = 30°, ux = 40 × cos 30° = 20√3 m/s, uy = 40 × sin 30° = 20 m/s
Time taken to reach the maximum height is given by,

Height above the tower is given by,

Therefore, the maximum height is, Hmax = H1 + H2 = 60 + 20 = 80 m
Now, the time taken to free fall from maximum height is, 
Thus, the total time is taken during the entire flight is given by, Ttotal = T1 + T2 = 2 + 4 = 6 s

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