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Test: Projectile Motion- 2 - EmSAT Achieve MCQ


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10 Questions MCQ Test - Test: Projectile Motion- 2

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Test: Projectile Motion- 2 - Question 1

If a projectile is thrown with velocify u at an angle θ with the horizontal, then the velocity at maximum height during the projectile's motion will be:

Detailed Solution for Test: Projectile Motion- 2 - Question 1

Projectile motion:

  • A kind of motion that is experienced by an object when it is projected near the Earth's surface and it moves along a curved path under the action of gravitational force.
  • Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • In case of a stone thrown in any direction, a stone thrown horizontally from a building, and a bullet fired from a gun are subjected to only gravitional force. But in case of a car moving in a straight line is a rectilinear type of motion. 

  • The velocity at maximum height of a projectile depends on the initial velocity and the angle at which it was launched. At maximum height, the vertical component of velocity becomes zero and the projectile is momentarily at rest before it starts falling down.
  • We can use the equations of motion to find the velocity at maximum height of the projectile. Let's assume that the projectile is thrown from the origin (0,0) and that the acceleration due to gravity is -g in the y-direction. The horizontal velocity remains constant during the motion of the projectile and is given by:

where u is the initial velocity and θ is the angle of projection.

The vertical velocity, Vy, at any point during the motion of the projectile is given by:

where t is the time elapsed since the projectile was launched.

At maximum height, the vertical velocity becomes zero. So, we can equate Vy to zero and solve for the time taken to reach maximum height:

Substituting this value of t in the expression for Vy, we get:

So, at maximum height, the velocity of the projectile is only the horizontal velocity:

vmax = vx = u cos(θ)

Therefore, the velocity at maximum height during the projectile's motion is u cos(θ).

Test: Projectile Motion- 2 - Question 2

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

Detailed Solution for Test: Projectile Motion- 2 - Question 2

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 + BXis 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

2) Horizontal range:

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

3) Height:

It is the maximum vertical distance traveled by the projectile.

 

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Test: Projectile Motion- 2 - Question 3

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

Detailed Solution for Test: Projectile Motion- 2 - Question 3

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- 2 - Question 4

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- 2 - Question 4

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- 2 - Question 5

Which of the following is NOT a projectile motion?

Detailed Solution for Test: Projectile Motion- 2 - Question 5

Projectile motion:

  • A kind of motion that is experienced by an object when it is projected near the Earth's surface and it moves along a curved path under the action of gravitational force.
  • Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • In case of a stone thrown in any direction, a stone thrown horizontally from a building, and a bullet fired from a gun are subjected to only gravitational force. But in case of a car moving in a straight line is a rectilinear type of motion. 

Test: Projectile Motion- 2 - Question 6

Two bodies A and, B are projected upwards such that the velocity of A is double the velocity of B. Then, the ratio of heights to which body A and body B will rise is

Detailed Solution for Test: Projectile Motion- 2 - Question 6

Maximum height: it the maximum height from point of projection that the projectile can reach.

Let, VA = Velocity of body A

VB = Velocity of the ball B

Given VB = u and VA = 2u

The maximum height reached by a body is given by, 

Or H ∝ u2

Test: Projectile Motion- 2 - Question 7

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- 2 - Question 7

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.

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

Where h is the maximum height.

Given that u = Vo m/s, θ = 30°

Maximum Height 

Test: Projectile Motion- 2 - Question 8

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- 2 - Question 8

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.

u= 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 -

 

Test: Projectile Motion- 2 - Question 9

shell is fired from a cannon with a speed ‘v’ at an angle θ with the horizontal direction as shown in the figure. At the highest point in its path, it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed of other piece immediately after the explosion is

Detailed Solution for Test: Projectile Motion- 2 - Question 9

Initial momentum =

Final momentum =

Hence conserving momentum

∴ v’ = 3 v cos θ

Test: Projectile Motion- 2 - Question 10

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- 2 - Question 10

Concept:

At the highest point of the projectile:

For projectile motion we know: a= 0 and a= -g

Diagram

Explanation:

V= 0 (at highest point)

usin 2θ = 2gH

u sinθ = √2gH

V= uy+ ayt

0 = √2gH - gt

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

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