Projectile Motion MCQ Quiz - Objective Question with Answer for Projectile Motion - Download Free PDF

Explanation:

(a) Correct: Due to horizontal acceleration caused by the wind and vertical acceleration due to gravity, the ball travels along a straight path.

(b) Incorrect: The simultaneous horizontal and vertical accelerations make the ball’s path straight.

(c) Correct: The time to reach the ground depends only on vertical motion. Using the equation h = (1/2) g t2:

98 = (1/2) × 9.8 × t2

t2 = 20

t = √20 ≈ 4.47 seconds

(d) Correct: Because the ball moves horizontally during its fall, the total distance it travels is greater than the vertical height of 98 meters.

Final Answer: (a), (c), (d)

Explanation:

The correct answer is (a).

• In projectile motion, the only force acting on the object (neglecting air resistance) is the gravitational force, which causes a constant acceleration downward.
• This acceleration remains constant in both magnitude and direction (vertically downward) throughout the motion, including at the highest point.
• At the highest point, the vertical component of the velocity becomes zero, but the horizontal velocity remains constant. However, acceleration due to gravity is still acting downward.
• Option (b) is incorrect because the acceleration is present even at the highest point, though the vertical velocity is momentarily zero.
• Option (c) is incorrect because acceleration due to gravity is always vertically downward and can be perpendicular to the instantaneous velocity, such as at the highest point where velocity is purely horizontal.
• Hence, option (a) is the correct answer.

Final Answer: (a)

Calculation:

For projection angle α, the range is:

R₁ = u² sin(2α) / g

For projection angle β, the range is:

R₂ = u² sin(2β) / g

Given that α + β = 90°, we can express β as:

β = 90° - α

⇒R₂ = u² sin(2(90° - α)) / g

Using the identity sin(180° - x) = sin(x), we get:

⇒R₂ = u² sin(2α) / g

Thus, R₁ = R₂, so the ratio is:

⇒ R₁ / R₂ = (u² sin(2α) / g) / (u² sin(2α) / g) = 1

Final Answer: R₁ / R₂ = 1

Calculation:

Equation of motion: F = m d(v)/dt

⇒-C(v_x⁠î + v_y⁠ĵ) - mgĵ = m dv_x ⁄ dt î + m dv_y ⁄ dt ĵ

⇒ -Cv_x = m dv_x ⁄ dt

⇒ m dv_x / dx × dx ⁄ dt = -C v_x

⇒ m dv_x = -C dx

⇒ v_x = u_x - Cx ⁄ m

Integrating bodth side

⇒ ∫x dx ⁄ (u_x - (C/m)x) = ∫t dt

⇒ u_x (1 - e^-Ct/m) = C⁄m x

⇒ x = m u_x⁄C [1 - e^-Ct/m]

⇒ x = 270 (1 - 1/e) = 270 (1 - 1/2.7)

⇒x = 170 m

Concept: 

In the projectile Motion, The Maximum Height is given by 

\(H~=~\frac{V^2sin^2~θ}{2g}\)

where, V = Velocity of Object, H = Maximum Height, θ = Angle of the velocity of an object with respect to ground.

Linear Momentum is given by,

Momentum = Mass × Velocity

Also, Angular Momentum is given by,

Angular Momentum L = Momentum × Perpendicular distance

​Calculation:

Given:

The particle of mass m and velocity v making an angle of 45° with horizontal as shown in the figure,

So at the Highest point, Velocity = Vcos 45 = \(\frac{V}{\sqrt2 }\)

Momentum = Mass× Velocity

Momentum = m× \(\frac{V}{\sqrt2 }\)

Also, Angular Momentum L = Momentum× Perpendicular distance

\(L~=~\frac{mv}{\sqrt2}\times{H}\)

where, H = Maximum Height when θ = 45° 

\(H~=~\frac{V^2sin^2~θ}{2g}~=~\frac{V^2sin^2~45}{2g}~=~\frac{V^2}{4g}\)

\(L~=~\frac{mV}{\sqrt2}\times\frac{V^2}{4g}\)

\(L~=~\frac{mV^3}{{4\sqrt2}{g}}\)

Hence it is proved that Angular Momentum is \(L~=~\frac{mV^3}{{4\sqrt2}{g}}\)  when θ = 45°.

CONCEPT:

• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.
• There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

EXPLANATION:

v = 30 m/s, u = 10 m/s, a = 4 m/s²

We know that,

⇒ v² = u² + 2aS

⇒ 2aS = v² – u²

⇒ 2 × 4 × S = 900 - 100

⇒ 8S = 800

CONCEPT:

Equation of Kinematics:

• These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

⇒ V = U + at

\(⇒ s = ut +\frac{1}{2}at^2 \)

⇒ V2 = U2+ 2as

Where, U = Initial velocity, V = Final velocity, g = Acceleration due to gravity, t = time, and h = height/Distance covered.

Velocity:

• The velocity of the particle is defined as the rate of change of its displacement and can be expressed as:

\(\vec v = \frac{{\overrightarrow {dx} }}{{dt}}\)

Acceleration:

• The acceleration of the particle is defined as the rate of change of its velocity and can be expressed as:

\(\vec a = \frac{{\overrightarrow {dv} }}{{dt}}\)

CALCULATION:

Given:

The ball is thrown vertically up with a speed (u) of 40 m/s.

Height of the tower (s) = - 240 m

Gravitational acceleration (a) = -10 m/s²

By using the second equation of motion,

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

Put values of u, s, and a in the second equation of motion.

We get,

\(⇒ -240 =(40\times t)+\left ( \frac{1}{2}\times -10\times t^{2} \right )\)

⇒ - 240 = 40t - 5t² 

• Here, the ball is going upward direction from point B to C saying positive and comes back in a downward direction from point C to D saying negative. Therefore, cancel out both distances BC and CD because of BC = -CD. Again the ball comes back to a downward direction from point D to E saying negative. So, the total distance traveled by the ball is -240 m.
• When the ball goes upward against gravity, the displacement is considered positive and gravity is negative. Now, when the ball comes downward, the displacement is considered negative and gravity is positive.

⇒ 5t² - 40t - 240 = 0

⇒ t² - 8t - 48 = 0

By solving the equation,

We get, t = 12 sec and t = - 4 sec. Taking positive value of time.

⇒ t = 12 sec

Alternate Method 

CALCULATION:

Given:

The ball is thrown vertically up with a speed (u_i) of 40 m/s.

Height of the tower (s) = 240 m

Gravitational acceleration (a) = 10 m/s²

When the ball throws from the top of the tower, it will travel the distance from point B to E, i.e.

BE = BC + CD + DE

The total time required for the displacement BE will be T = t1 + t2

Where, t₁ = time for BC + CD and t₂ = time for DE

The time required for distance BC + CD:

\(t_1 = \frac{2u_i}{g}\)

\(t_1 = \frac{2 \times 40}{10}\)

t₁ = 8 sec

The time required for distance DE:

The distance travel by the ball is 240 m with the velocity of 40 m/s having the acceleration due to gravity is 10 m/s².

\(⇒ s = ut +\frac{1}{2}at^2 \)

\(⇒ 240 = 40 \times t_2 +\frac{1}{2}10\times t^2 _2\)

\(⇒ 240 = 40 t_2 +5 t^2 _2\)

\(⇒ 5 t^2 _2+40 t_2 -240=0\)

t₂ = 4 sec

So, the total time required for the displacement, T = t₁ + t₂

T = 8 + 4 = 12 sec

CONCEPT:

• The angle of projection: The angle between the initial velocity of a body from a horizontal plane through which the body is thrown, is known as the angle of projection.
  • The angle of projection for which the maximum height of the projectile is equal to the horizontal range has to be determined.
  • A body can be projected in two ways :
    1. Horizontal projection-When the body is given an initial velocity in the horizontal direction only.
    2. Angular projection-When the body is thrown with an initial velocity at an angle to the horizontal direction.
• Maximum height of projectile: It is when the projectile reaches zero vertical velocity is called the maximum height.
  • From this point, the vertical component of the velocity vector will point downwards.
  • The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object.

FORMULA:

\(H = \frac{{\mathop V\nolimits_O^2 {{{\mathop{\rm Sin}\nolimits} }^2}θ }}{{2g}}\)

where, H is the maximum height, v_o = initial velocity, g = acceleration due to gravity, θ = is the angle of the initial velocity from the horizontal plane (radians or degrees).

CALCULATION:

Given that, v = u/2

At maximum height, the vertical velocity component ceases and only the horizontal component is present

i.e. v = u cosθ

According to the given problem, v = u/2

\(\therefore \frac{u}{2} = u\cos θ \to \cos θ = \frac{1}{2}\)

θ = 60° 

The correct option is 60°.

Concept:

The angle of projection in horizontal range is the maximum interval in horizontal distance.

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

The height of the maximum vertical interval covered by the projectile.

\({\rm{h}} = \frac{{{{\rm{v}}^2}{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}}{{2{\rm{g}}}}\)

Calculation:

The two particles are projected from same point with same speed, same range and different height.

Let the two particles be p₁ and p₂.

So, the angle of projection of particle p₁ is θ and particle p₂ is 90 - θ

The range of the projectile motion is given by the formula:

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

∵ [sin 2θ = 2 sin θ cos θ]

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

The height of the projectile motion is given by the formula:

 \({\rm{h}} = \frac{{{{\rm{v}}^2}{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}}{{2{\rm{g}}}}\)

Particle p₁: (Speed ‘u’)

Now, the range of the projectile motion of particle p₁ is:

\( \Rightarrow {{\rm{R}}_1} = \frac{{2{{\rm{u}}^2}{\rm{sin\theta cos\theta }}}}{{\rm{g}}}\)

The height of the projectile motion of particle p₁ is:

\( \Rightarrow {{\rm{h}}_1} = \frac{{{{\rm{u}}^2}{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}}{{2{\rm{g}}}}\)

\( \Rightarrow 2{{\rm{h}}_1} = \frac{{{{\rm{u}}^2}{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}}{{\rm{g}}}\)           ----(1)

Particle p₂: (Speed ‘u’)

Now, the range of the projectile motion of particle p₂ is:

\( \Rightarrow {{\rm{R}}_2} = \frac{{2{{\rm{u}}^2}{\rm{sin\theta cos\theta }}}}{{\rm{g}}}\)

The height of the projectile motion of particle p₂ is:

\( \Rightarrow {{\rm{h}}_2} = \frac{{{{\rm{u}}^2}{\rm{si}}{{\rm{n}}^2}\left( {90 - {\rm{\theta }}} \right)}}{{2{\rm{g}}}}\)

\( \Rightarrow {{\rm{h}}_2} = \frac{{{{\rm{u}}^2}{{\cos }^2}{\rm{\theta }}}}{{2{\rm{g}}}}\)

\( \Rightarrow 2{{\rm{h}}_2} = \frac{{{{\rm{u}}^2}{{\cos }^2}{\rm{\theta }}}}{{\rm{g}}}\)           ----(2)

The range of both particle is same.

Now, from the options, R² is related h₁ and h₂.

So,

\( \Rightarrow {{\rm{R}}^2} = \frac{{4{{\rm{u}}^4}{{\sin }^2}{\rm{\theta }}{{\cos }^2}{\rm{\theta }}}}{{{{\rm{g}}^2}}}\)

\( \Rightarrow {{\rm{R}}^2} = 4\left( {\frac{{{{\rm{u}}^2}{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}}{{\rm{g}}}} \right) \times \left( {\frac{{{{\rm{u}}^2}{{\cos }^2}{\rm{\theta }}}}{{\rm{g}}}} \right)\)

On substituting equation (1) and (2) in above equation,

⇒ R² = 4(2h₁)(2h₂)

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.

EXPLANATION:

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

\(\Rightarrow {{\rm{T}}_{1/2}} = \frac{{{\rm{v\;sin\theta }}}}{{\rm{g}}}\)

Where T_1/2 = 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.

\(\Rightarrow {\rm{T}} = \frac{{2{\rm{\;v\;sin\theta }}}}{{\rm{g}}}\)

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.

\(\Rightarrow R = \frac{{{v^2}sin2\theta }}{g}\)

Where R is the total distance covered by the projectile.

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

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

CONCEPT:

• When an object is thrown at an angle θ with some initial velocity, it goes in projectile motion before hitting the ground.

Horizontal Range in projectile motion is given by:

R = \( \frac{u^2 sin2θ}{g} \)

Where u is initial velocity θ is an angle of projection with horizontal and g is the gravitational acceleration.

• Max value of sin α is '1' at α = 90° 

EXPLANATION:

• Horizontal Range R = \( \frac{u^2 sin2θ}{g} \) will change according to the speed and angle of projection θ. 
• In question, It is Given speed is the same for all. So, Range will change according to θ.

For range R = \( \frac{u^2 sin2θ}{g} \) to be maximum, sin 2θ should be maximum.

sin 2θ = 1 ⇒ sin 2θ = sin 90° 

2θ = 90

θ = 45° 

So, at an angle of 45°, the stone will cover maximum horizontally.

• The answer is option 2 i.e 45°

• For θ = 45° horizontal Range is always maximum.
• The maximum value of Range will be  R_max = \( \frac{u^2}{g} \)

The correct answer is Acceleration is constant.

Concept:

•  The rate of change of velocity of a body is called the acceleration of that body.
• It is a vector quantity that has both magnitudes as well as direction. 

\(acceleration\left( a \right) = \frac{{Change\;in\;velocity}}{{time\;taken}} = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}or\frac{{dv}}{{dt}}\)

• The slope (m) of the displacement-time graph at any point will give the instantaneous velocity at that point.
• The slope of any graph at a point is given by: Slope (m) = tanθ
• Where θ is the angle made by the point with the x-axis.

Explanation:

• The given graph shows acceleration vs time graph
• In our case, the slope of the given graph is zero since the angle (θ) in our case is zero
• i.e., \(\tan \theta = \frac{{da}}{{dt}} \Rightarrow \frac{{da}}{{dt}} = 0\)
• From this we can see that rate of change of acceleration is zero hence the acceleration throughout the motion must remain constant.
• Hence option 2 is correct

CONCEPT:

Projectile motion

• If an object is given an initial velocity in any direction and then allowed to travel freely under gravity, then the object is called a projectile and the motion of the projectile is called the projectile motion.
• There is no force other than the gravity acts on the projectile during the flight.

EXPLANATION:

• When an object is projected at an angle from the surface of the earth, then there will be two components of the velocity.
• One component of velocity will act along the horizontal direction and the other will act in the vertical direction.
• There is no force other than the gravity acts on the projectile during the flight. So only one acceleration due to gravity in the vertically downward direction will act on the object.
• Since there is no acceleration in the horizontal direction so the horizontal velocity will remain constant during the whole flight.
• At the highest point, the vertical velocity of the object will be zero and the whole velocity of the particle will be along the horizontal direction.
• The acceleration on the particle during the flight will be due to gravity, so the direction of acceleration will be vertical downward at the highest point.
• So at the highest point, the direction of velocity is horizontal and the direction of acceleration is vertically downward, so the angle between the direction of velocity and acceleration at the highest point of a projectile path is 90°. Hence, option 3 is correct.

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.

Explanation:

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

The mathematical expression of the horizontal range is -

\(⇒ H = \frac{{{v^2}{{\sin }^2}\theta }}{{2g}}\)      ------- (1)

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.

\(⇒ {{T}} = \frac{{2{{v\;sin\theta }}}}{{{g}}}\)  

On squaring both, we get

\(⇒ {{T^2}} = \frac{{4{{\;v^2\;sin^2\theta }}}}{{{g^2}}}\)     ------ (2)

On dividing equations 1 and 2, we get

\(⇒ \frac{H}{T^2} = \frac{\frac{{{v^2}{{\sin }^2}\theta }}{{2g}}}{\frac{{{4v^2}{{\sin }^2}\theta }}{{g^2}}}=\frac{g}{8}\)

\(\Rightarrow H =\frac{gT^2}{8}\)

Important Points ​

Time is taken to reach maximum height: 

it is half of the total time of flight.

\(⇒ {{\rm{T}}_{1/2}} = \frac{{{\rm{v\;sin\theta }}}}{{\rm{g}}}\)

Range: 

The range of the motion is fixed by the condition y = 0.

\(⇒ R = \frac{{{v^2}sin2\theta }}{g}\)

Where R is the total distance covered by the projectile.

Concept: 

• 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.
• When a particle moves in projectile motion, its velocity has two components.
  1. ​​vertical component (u sinθ)
  2. horizontal component(u cosθ)

• 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.

• The range of flight is given by

\(R = \frac{u^2sin2θ}{g}\)

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

• The maximum height a projectile can attain 

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

Calculation:

Let us consider the initial velocity of the projectile is u m/s, H is the maximum height of the projectile and R is its horizontal range.

• The range of flight is given by

\(R = \frac{u^2sin2θ}{g}\)

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

• The maximum height a projectile can attain 

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

To find θ, let us divide the maximum height of the projectile by horizontal range

\(\frac{H}{R} = \left( {\frac{{\frac{{{u^2}{{\sin }^2}θ }}{{2g}}}}{{\frac{{{u^2}\sin 2θ }}{g}}}} \right)\)

\(\frac{H}{R} = \left( {\frac{{{u^2}{{\sin }^2}θ \times g}}{{{u^2}\sin 2θ \times 2g}}} \right)\)

By solving the above equations; we get

\(\frac{H}{R} = \frac{{{{\sin }^2}θ }}{{2\sin 2θ }}\)       ( we know that sin 2θ = 2 sinθ cosθ )

\(\frac{H}{R} = \frac{{\sin θ \sin θ }}{{2 \times 2\sin θ \cos θ }}\)

\(\frac{H}{R} = \frac{{\sin θ }}{{4\cos θ }}\)     (Since cosθ/sinθ = cotθ)

⇒ R = 4 H cot θ