This tutorial discusses projectile motion with practice questions(MCQ’s) and answers.

Each question and answer is given in such a way that the basic to advanced concepts of projectile motion are clear.

Of course, each question is given in quiz form so you can practice on your own.

What is projectile motion?

- Motion of an object in a straight line
- Motion of an object in a circular path
- Motion of an object thrown or launched into the air
- Motion of an object in a zigzag pattern

## Explanation

Projectile motion is the motion of an object that is thrown or launched into the air and moves along a curved path under the influence of gravity.

It is a type of motion in two dimensions and is characterized by its horizontal and vertical components.

What is the vertical component of velocity during the entire flight of a projectile?

- It remains constant
- It increases
- It decreases
- It changes due to the acceleration of gravity

## Explanation

Vertical component of velocity changes due to the acceleration of gravity, which is constant and acts downward on the projectile.

As a result, the vertical component of velocity decreases until it reaches zero at the highest point of the trajectory and then increases in the opposite direction until it reaches the initial velocity at the end of the trajectory.

What is the horizontal component of velocity during the entire flight of a projectile?

- It remains constant
- It increases
- It decreases
- It changes due to the acceleration of gravity

## Explanation

Horizontal component of velocity remains constant during the entire flight of a projectile, as there is no force acting in the horizontal direction to change its magnitude or direction.

This means that the projectile moves with a constant horizontal speed, which is independent of its vertical motion.

What is the path of a projectile?

- A straight line
- A parabolic curve
- A circle
- An ellipse

## Explanation

Path of a projectile is a parabolic curve, which is a result of its motion in two dimensions under the influence of gravity.

Horizontal component of motion is uniform, while the vertical component of motion is accelerated due to the force of gravity acting in the vertical direction.

As a result, the projectile follows a curved path, which is a parabolic curve.

What is the maximum height reached by a projectile?

- Initial height of the projectile
- Final height of the projectile
- Highest point of the trajectory
- Lowest point of the trajectory

## Explanation

Maximum height reached by a projectile is the highest point of the trajectory, where the vertical component of velocity is zero.

At this point, the projectile stops moving upward and begins to fall downward due to the force of gravity.

Maximum height reached by the projectile depends on the initial velocity and angle of projection.

What is the range of a projectile?

- Initial horizontal displacement of the projectile
- Final horizontal displacement of the projectile
- Horizontal distance traveled by the projectile
- Horizontal distance between launch point and landing point

## Explanation

Range of a projectile is the horizontal distance between launch point and landing point, which is determined by the initial velocity and angle of projection.

Horizontal component of velocity remains constant during the entire flight of the projectile, and the range depends on the time of flight and the horizontal velocity of the projectile.

What is the time of flight of a projectile?

- Time it takes for the projectile to reach its maximum height
- Time it takes for the projectile to reach its final height
- Time it takes for the projectile to travel a certain distance horizontally
- Time it takes for the projectile to return to its initial height

## Explanation

Time of flight of a projectile is the time it takes for the projectile to return to its initial height, which is equal to twice the time it takes for the projectile to reach its maximum height.

Time of flight is determined by the initial velocity and angle of projection, and it is independent of the mass of the projectile.

What is the angle of projection for maximum range of a projectile?

- 30 degrees
- 45 degrees
- 90 degrees
- 180 degrees

## Explanation

Angle of projection for maximum range of a projectile is 45 degrees. However, when air resistance is neglected, the maximum range is achieved when the angle of projection is 90 degrees, which corresponds to a vertical projection.

In this case, Horizontal component of velocity remains constant, and projectile travels maximum distance horizontally before returning to the ground.

What is the angle of projection for height and range are equal of a projectile?

- tan
^{-1}(1/4) degrees - tan
^{-1}(4) degrees - 45 degrees
- (π/3) degrees

## Explanation

The angle of projection for maximum height of a projectile is 45 degrees.

Angle results in the maximum height of the projectile because it allows for the largest vertical component of velocity, which is necessary to overcome the acceleration due to gravity and reach the maximum height.

At this angle, the horizontal component of velocity is also sufficient to ensure a long flight time, allowing the projectile to reach a greater height.

What happens to the time of flight of a projectile if the initial velocity is doubled?

- It is doubled
- It is halved
- It remains the same
- It is quadrupled

## Explanation

The time of flight of a projectile is direct proportional to the vertical component of velocity.

What happens to the maximum height of a projectile if the angle of projection is increased?

- Maximum height increases
- Maximum height decreases
- Maximum height remains the same
- It depends on the initial velocity of the projectile

## Explanation

As the angle of projection is increased, the vertical component of velocity decreases, resulting in a lower maximum height. This is because the projectile spends less time in the air and has a lower upward velocity component, so it doesn’t reach the same height as it would at a lower angle of projection with a larger upward velocity component.

What is the formula for the range of a projectile?

- R = v
_{0}sin(θ) / g - R = v
_{0}cos(θ) / g - R = v
_{0}^{2}sin(2θ) / g - R = v
_{0}^{2}cos(2θ) / g

## Explanation

The formula for the range of a projectile is R = v_{0}^{2} sin(2θ) / g, where R is the range, v_{0} is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. This formula assumes that air resistance is negligible.

What is the formula for the maximum height of a projectile?

- h = v
_{0}/ g - h = v
_{0}^{2}sin^{2}(θ) / 2g - h = v
_{0}^{2}cos^{2}(θ) / 2g - h = v
_{0}^{2}sin(2θ) / g

## Explanation

Formula for the maximum height of a projectile is h = v_{0}^{2} sin^{2}(θ) / 2g, where h is the maximum height, v_{0} is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

This formula assumes that air resistance is negligible.

What happens to the range of a projectile if the initial velocity is doubled?

- Range is doubled
- Range is halved
- Range remains the same
- Range is quadrupled

## Explanation

The range of a projectile is proportional to the product of the initial velocity and the time of flight.

Doubling the initial velocity will therefore result in a doubling.

What is the acceleration of a projectile at the highest point of its trajectory?

- Zero
- -9.81 m/s
^{2} - 9.81 m/s
^{2} - Depends on the initial velocity of the projectile

## Explanation

At the highest point of its trajectory, the vertical velocity of the projectile is zero, and the only force acting on the projectile is gravity.

Therefore, the acceleration of the projectile at the highest point is equal to the acceleration due to gravity, which is -9.81 m/s^{2} (assuming no air resistance).

Which is correct option below?

- V
_{x}parallel to the ground. - V
_{x}perpendicular to the ground. - V
_{x}in the direction of acceleration. - There is no horizontal component of velocity in projectile motion.

## Explanation

The horizontal component of a projectile’s velocity is the component of velocity parallel to the ground. This velocity remains constant throughout the projectile’s motion since there is no horizontal acceleration in projectile motion (assuming no air resistance).

Which is correct option below?

- Motion in two dimensions due to influence of gravity.
- Motion in one dimension due to influence of gravity.
- Motion in three dimensions due to influence of gravity.
- Motion in two dimensions due to influence of electromagnetic forces.

## Explanation

Projectile motion is the motion of an object in two dimensions (x and y) due to the influence of gravity.

Object is typically thrown or launched at an angle to the horizontal and follows a parabolic path.

What is the vertical component of a projectile’s velocity at the highest point of its trajectory?

- Zero
- Negative
- Positive
- Depends on the height of the projectile’s initial position

## Explanation

At the highest point of its trajectory, the vertical velocity of the projectile is zero. This is because the projectile has reached its maximum height and is about to start falling back down due to gravity.

Vertical velocity of the projectile changes sign at the highest point of its trajectory, going from positive (upward) to negative (downward).

What is the equation for the vertical displacement d(t) of a projectile as a function of time?

- d(t) = v
_{0}cosθt - d(t) = (v
_{0}sinθ)t – 1/2gt² - d(t) = v
_{0}sinθt - d(t) = (v
_{0}sinθ)t + 1/2gt²

## Explanation

Vertical displacement of a projectile as a function of time is given by the equation d = V₀sinθt – 1/2gt², where V₀ is the initial velocity of projectile, g is acceleration due to gravity, t is the time elapsed, and Negative sign indicates that acceleration is in downward direction.

What happens to the range of a projectile when the launch angle is increased from 30 degrees to 60 degrees?

- Range increases
- Range decreases
- Range stays the same
- Range is impossible to determine without more information

## Explanation

Range of a projectile is given by the equation R = V₀²sin2θ/g.

When the launch angle is increased from 30 degrees to 60 degrees, the value of sin2θ also increases by the same factor, but gravitational acceleration g remains constant.

Therefore, the range stays the same.

What happens to the time of flight of a projectile when the initial velocity is doubled?

- Time of flight increases by a factor of √2
- Time of flight decreases by a factor of √2
- Time of flight doubles
- Time of flight will be constant

## Explanation

The time of flight of a projectile is given by the equation t = 2V₀sinθ/g. When the initial velocity is doubled, the value of V₀ also doubles.

So, the time of flight increases by a factor of 2.

What happens to the maximum height reached by a projectile when the initial velocity is doubled?

- Maximum height reached quadruples
- Maximum height reached doubles
- Maximum height reached increases by a factor of √2
- Maximum height reached stays the same

## Explanation

Maximum height reached by a projectile is given by the equation h = V₀²sin²θ/2g.

When the initial velocity is doubled, the value of V₀ also doubles, so the maximum height reached increases by a factor of 4 (quadruples).

h’= 4h

What happens to the maximum height reached by a projectile when the launch angle is increased from 30 degrees to 45 degrees?

- Maximum height reached increases
- Maximum height reached decreases
- Maximum height reached stays the same
- Maximum height reached is impossible to determine without more information

## Explanation

Maximum height reached by a projectile is given by the equation h = V₀²sin²θ/2g.

When the launch angle is increased from 30 degrees to 45 degrees, the value of sin²θ increases by same factor that the value of sinθ increases.

What is equation for time of flight of a projectile?

- t = V₀sinθ/g
- t = V₀cosθ/g
- t = 2V₀sinθ/g
- t = 2V₀cosθ/g

## Explanation:

Time of flight of a projectile is given by the equation t = 2V₀sinθ/g, where V₀ is the initial velocity of the projectile, θ is the launch angle, and g is the acceleration due to gravity.

This equation assumes no air resistance and a level launch surface.

A particle is launched from a certain height h_{0} with an initial velocity u_{0} at an angle θ. What will be the maximum height in this case?

- h
_{max}= h_{0}+ u_{0}^{2}sin^{2}(θ) / 2g - h
_{max}= u_{0}^{2}sin^{2}(θ) / 2g - h
_{max}= h_{0}+ u_{0}^{2}sin^{2}(θ) / g - h
_{max}= h_{0}– u_{0}^{2}sin^{2}(θ) / 2g

## Explanation

Formula for the maximum height of a projectile is h = v_{0}^{2} sin^{2}(θ) / 2g, where h is the maximum height, v_{0} is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

And if projecting from specified height h_{0}, then the intial height will be added. So, h_{0} + u_{0}^{2} sin^{2}(θ) / 2g