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Basics Of Projectile Motion

Learn formula to calculate projectile motion time of flight, height of a projectile, and range of projectile. Practice projectile motion problems with solutions and examples.

Projectile Motion from Ground to Ground

  • A 2-D motion can be analyzed as a combination of two independent motions in x and y directions with accelerations \(a_x\) and \(a_y\) respectively.
  • These independent motions can be expressed as:
  1. Motion of a particle under constant velocity, with zero acceleration in horizontal (x) direction i.e., \(a_x\)= 0.
  2. Motion of a particle under constant acceleration in vertical (y) direction i.e., \(a_y\,=\,-g\).

Illustration Questions

A particle is projected at an angle of 37° with the horizontal, has a velocity of 5 m/s. Calculate the horizontal and the vertical components of initial velocity. \(\left ( cos\,37°=\dfrac {4}{5}, \,\,\,sin\,37°=\dfrac {3}{5} \right)\)

A \(v_x=4\,m/s\), \(v_y\)\( = 3 \,m/s\)

B \(v_x\)\(= 5\, m/s\), \(v_y\) \( = 8 \,m/s\)

C \(v_x\) \(= 6 \,m/s\), \(v_y\)\(= 4\, m/s\)

D \(v_x\)\(= 7 \,m/s\), \(v_y\) \( = 3\, m/s\)

×

Horizontal component of velocity

\(v_x\)\(v\,cos\,\theta\)

\(v_x\) = 5 × cos 37°

= 4  m/s

image image

Vertical component of velocity

\(v_y\)\(v\,sin\,\theta\)

\(v_y\) = 5 × sin 37°

= 3 m/s

image image

A particle is projected at an angle of 37° with the horizontal, has a velocity of 5 m/s. Calculate the horizontal and the vertical components of initial velocity. \(\left ( cos\,37°=\dfrac {4}{5}, \,\,\,sin\,37°=\dfrac {3}{5} \right)\)

A

\(v_x=4\,m/s\), \(v_y\)\( = 3 \,m/s\)

.

B

\(v_x\)\(= 5\, m/s\)\(v_y\) \( = 8 \,m/s\)

C

\(v_x\) \(= 6 \,m/s\), \(v_y\)\(= 4\, m/s\)

D

\(v_x\)\(= 7 \,m/s\)\(v_y\) \( = 3\, m/s\)

Option A is Correct

Time of Flight (From Ground to Ground)

  • Consider a projectile motion of a ball as shown in the figure.
  • The time taken by the ball to reach from 0 to A or the time duration in which the ball remains in air is known as 'Time of Flight'.
  • The time of flight is calculated by considering the motion only along Y-axis.
  • When particle reaches at point 'A' the displacement along Y-axis becomes zero.
  • Thus,  \(\Delta y = 0\)

Also,  \(u_y=u\;sin\,\theta\) and \(a_y=-g\)

From \(\Delta y = u_yt+\dfrac {1}{2}a_yt^2\)

\(\Rightarrow0=u\,sin\,\theta \,t-\dfrac {1}{2}gt^2\)

\(\Rightarrow t=\dfrac {2u\,sin\,\theta}{g}\)

Time of flight

\(T=\dfrac {2u\,sin\,\theta}{g}=\dfrac {2u_y}{g}\)     

Illustration Questions

A particle is projected with a velocity of \(v = 25 \,m/s\) at an angle of \(\theta=\)\(37°\) with the horizontal. Calculate the time of flight. (\(g = 10 \,m/s^2, \,\,sin \,37° =\) \(\dfrac {3}{5}\))

A 2 sec

B 3 sec

C 4 sec

D 5 sec

×

Velocity of particle, v = 25 m/s

Angle of projection, \(\theta =\)37°

image

Time of Flight

\(T=\dfrac {2v_y}{g}=\dfrac {2v\,sin\theta}{g}\)

\(T=\dfrac {2(25)\;sin\,37°}{10}\)

= 3 sec

image image

A particle is projected with a velocity of \(v = 25 \,m/s\) at an angle of \(\theta=\)\(37°\) with the horizontal. Calculate the time of flight. (\(g = 10 \,m/s^2, \,\,sin \,37° =\) \(\dfrac {3}{5}\))

A

2 sec

.

B

3 sec

C

4 sec

D

5 sec

Option B is Correct

Maximum Height Attained by a Particle

  • In projectile motion, the horizontal and vertical components of a particle are mutually independent to each other.
  • So, height attained by a particle, will also be independent of its motion along X-axis. 
  • Parameters along Y-axis are:-
  • Initial velocity \(u_y=u\,sin\theta\) 

    \(a_y=-g\)

  • At maximum height, vertical component of velocity becomes zero.
  • Using equation of motion-
  • \(v^2=u^2+2as\)

    \((0)^2=u^2\,sin^2\theta-2g(H_{max})\)

    \(H_{max}=\dfrac {u^2\,sin^2\theta}{2g}\)

     

Illustration Questions

A particle is projected with an initial velocity \( u = 25 \,m/s\) at an angle of \(\theta =\)\(37°\) with the horizontal. Calculate the maximum height attained by the particle. (\(g = 10\, m/s^2, \,\,\,sin \,37° = \dfrac{3}{5}\))

A \(40 \,m\)

B \(30 \,m\)

C \(\dfrac {45}{4}m\)

D \(20 \,m\)

×

Velocity of particle, u = 25 m/s

Angle of projection, \(\theta =\)37°

image

Maximum height attained by particle

\(H_{max}=\dfrac {(u_y)^2}{2g}=\dfrac {u^2\,sin^2\theta}{2g}\)

\(=\dfrac {(25)^2(sin37°)^2}{2(10)}=\dfrac {45}{4}\,m\)

image image

A particle is projected with an initial velocity \( u = 25 \,m/s\) at an angle of \(\theta =\)\(37°\) with the horizontal. Calculate the maximum height attained by the particle. (\(g = 10\, m/s^2, \,\,\,sin \,37° = \dfrac{3}{5}\))

A

\(40 \,m\)

.

B

\(30 \,m\)

C

\(\dfrac {45}{4}m\)

D

\(20 \,m\)

Option C is Correct

Resolving the Velocity Along and Perpendicular to the Inclined Plane

  • The case in which a particle is projected on an inclined plane, at an angle \(\theta\), the selection of proper axes is required.
  • For simplicity, X-axis is taken along the inclined plane and Y-axis is taken perpendicular to the inclined plane.

  • X-component of velocity

\(u_x=u\,cos(\theta-\phi)\)

  • Y-component of velocity

\(u_y=u\,sin(\theta-\phi)\)

where \('\phi'\) is the angle at which the plane is inclined

Illustration Questions

A particle is projected on an inclined plane as shown in the figure. Calculate the X and Y components of initial velocity. Given  \(u = 12 \,m/sec\),  \(\theta=\)\(75°\),  \(\phi=\)\(15°\)

A \(u_x=\)\(3\, m/s\),  \(u_y=\) \(7\, m/s\)

B \(u_x=\) \(2 \,m/s\),  \(u_y=\) \(6\, m/s\)

C \(u_x=\)\( 8 \,m/s\),  \(u_y=\) \(9 \,m/s\)

D \(u_x=\) \(6 \,m/s\),  \(u_y=\) \(6\sqrt 3\) \(m/s\)

×

Velocity of particle, \( u = 12 \,m/sec\) 

Angle of projection, \(\theta =\)\(75°\)

Angle of inclination, \(\phi=\)\(15°\)

X-component of initial velocity

\(u_x=u\,cos(\theta-\phi)\)

\(=12 \,cos (75° – 15° )\)

\(= 12 ×\) \(\dfrac {1}{2}\)

= 6 m/s

image

Y-component of initial velocity

\(u_y=u\,sin(\theta-\phi)\)

\(=12 \,sin (75° – 15° ) \)

\(= 12 ×\) \(\dfrac {\sqrt 3}{2}\)

= \(6\sqrt 3\) \(m/s\)

image

A particle is projected on an inclined plane as shown in the figure. Calculate the X and Y components of initial velocity. Given  \(u = 12 \,m/sec\),  \(\theta=\)\(75°\),  \(\phi=\)\(15°\)

image
A

\(u_x=\)\(3\, m/s\),  \(u_y=\) \(7\, m/s\)

.

B

\(u_x=\) \(2 \,m/s\),  \(u_y=\) \(6\, m/s\)

C

\(u_x=\)\( 8 \,m/s\),  \(u_y=\) \(9 \,m/s\)

D

\(u_x=\) \(6 \,m/s\),  \(u_y=\) \(6\sqrt 3\) \(m/s\)

Option D is Correct

Dependence of Time of Flight on Vertical Component of Velocity

  • When a projectile is projected on a horizontal plane, then its horizontal displacement is called the 'range of projectile'.
  • To find range take the given 2-D motion as a combination of two independent 1-D motions i.e., one along the vertical direction and other along the horizontal direction.

  • Parameters along Y-axis are:-

Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and

acceleration, \(a_y=-g\)  when particle reaches at A. 

\(\Delta y=0\)

\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)

T = time of flight

\(\dfrac {2u\,sin\,\theta}{g}\)

  • Parameters along X-axis are:-

Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and

acceleration, \(a_x=0\)

\(T=\dfrac {2u\,sin\,\theta}{g}\)

\(R=u_x\cdot T\)

\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)

So, Range \(=\dfrac {2u\,sin\,\theta \cdot u\,cos\,\theta}{g} \)

\(R=\dfrac {2\,u_x\cdot u_y}{g}\)

or, Range \(\propto\) Product of two components of initial velocity

Illustration Questions

Here, X and Y-components of initial velocity for projectiles of 4 particles are given- (g=10 m/s2) P1 : uX = 300 m/s,  uY = 10 m/s P2 : uX = 200 m/s,  uY = 30 m/s P3 : uX = 10 m/s,  uY = 400 m/s P4 : uX = 20 m/s,  uY = 100 m/s Which projectile has maximum time of flight?

A P1 

B P2

C P3

D P4

×

Time of flight for particle 1

\(T_1=\dfrac {2u_{y_{(1)}}} {g}\)

\(=\dfrac {2(10)}{10}=2\,sec\)

image

Time of flight for particle 2

\(T_2=\dfrac {2u_{y_{(2)}}} {g}\)

\(=\dfrac {2(30)}{10}=6\,sec\)

image

Time of flight for particle 3

\(T_3=\dfrac {2u_{y_{(3)}}} {g}\)

\(=\dfrac {2(400)}{10}=80\,sec\)

image

Time of flight for particle 4

\(T_4=\dfrac {2u_{y_{(4)}}} {g}\)

\(=\dfrac {2(100)}{10}=20\,sec\)

image

Time of flight for particle C is maximum.

image

Here, X and Y-components of initial velocity for projectiles of 4 particles are given- (g=10 m/s2) P1 : uX = 300 m/s,  uY = 10 m/s P2 : uX = 200 m/s,  uY = 30 m/s P3 : uX = 10 m/s,  uY = 400 m/s P4 : uX = 20 m/s,  uY = 100 m/s Which projectile has maximum time of flight?

A

P1 

.

B

P2

C

P3

D

P4

Option C is Correct

Conditions for Maximum Range

  • When a projectile is projected on a horizontal plane, then its horizontal displacement is called the 'range of projectile'.
  • To find range take the given 2-D motion as a combination of two independent 1-D motions i.e., one along the vertical direction and other along the horizontal direction.

 

  • Parameters along Y-axis are:-

Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and

acceleration, \(a_y=-g\)  when particle reaches at A. 

\(\Delta y=0\)

\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)

T = time of flight

\(\dfrac {2u\,sin\,\theta}{g}\)

  • Parameters along X-axis are:-

Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and

acceleration, \(a_x=0\)

\(T=\dfrac {2u\,sin\,\theta}{g}\)

\(R=u_x\cdot T\)

\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)

  • Since, \(2\,sin\,\theta\;cos\,\theta=sin\,2\theta\)

\(\therefore\) Range (R) \(=\dfrac {u^2\,sin\,2\theta}{g} \)

For range to be maximum \(sin\,2\theta=1\)

\(\theta=\)45°

\(R_{max}=\dfrac {u^2}{g}\)

Illustration Questions

A particle is projected with a velocity of 25 m/s. Calculate the maximum range. (g = 10 m/s2)

A 62.5 m

B 65 m

C 64 m

D 70 m

×

Velocity of particle, u = 25 m/s

image

Range of particle,

\(R_{max}=\dfrac {(u)^2}{g}\)

\(R_{max}=\dfrac {(25)^2}{10}\)

\(=\dfrac {625}{10}=62.5\;m\)

image

A particle is projected with a velocity of 25 m/s. Calculate the maximum range. (g = 10 m/s2)

A

62.5 m

.

B

65 m

C

64 m

D

70 m

Option A is Correct

Projection of a Particle from Some Height

  • Consider a particle is projected from some height, H.
  • To calculate the time taken by the particle to reach the ground, select Y-axis along the height H and X-axis perpendicular to the height H, as shown in the figure.
  • For motion along Y-axis:

Initial velocity along Y-axis, \(u_y=0\)

\(\Delta y=-H\)

\(a_y=-g\)

as  \(\Delta y=u_yt+\dfrac {1}{2}a_yt^2 \)

\(0×t+\dfrac {1}{2}×(-g)(t^2)=-H\)

\(t=\sqrt {\dfrac {2H}{g}}\)     

Illustration Questions

A particle is projected horizontally from the top of a building with a velocity u = 5 m/sec. The height of the building is H = 20 m. Calculate the time taken by the particle to reach the ground. (g = 10 m/s2)

A 3 sec

B 2 sec

C 5 sec

D 6 sec

×

Vertical component of initial velocity, \(u_{y}=\) 0

Vertical component of acceleration, \(a_y=\) – g

image

Height of building, H = 20 m

image

Time taken by particle is 

\(t=\sqrt {\dfrac {2H}{g}}\)

\(t=\sqrt {\dfrac {2×(20)}{10}}\)

= 2 sec

image

A particle is projected horizontally from the top of a building with a velocity u = 5 m/sec. The height of the building is H = 20 m. Calculate the time taken by the particle to reach the ground. (g = 10 m/s2)

A

3 sec

.

B

2 sec

C

5 sec

D

6 sec

Option B is Correct

Range of Projectile

  • When a projectile is projected on a horizontal plane, then its horizontal displacement is called the 'range of projectile'.
  • To find range take the given 2-D motion as a combination of two independent 1-D motions i.e., one along the vertical direction and other along the horizontal direction.

  • Parameters along Y-axis are:-

 Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and

acceleration, \(a_y=-g\)  when particle reaches at A. 

\(\Delta y=0\)

\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)

T = time of flight

\(\dfrac {2u\,sin\,\theta}{g}\)

  • Parameters along X-axis are:-

Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and

acceleration, \(a_x=0\)

\(T=\dfrac {2u\,sin\,\theta}{g}\)

\(R=u_x\cdot T\)

\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)

 

  •  Range (R) \(=\dfrac {2u\,sin\theta \cdot u\,cos\theta}{g} \)

\(R=\dfrac {2\,u_x\cdot u_y}{g}\)

hence, we conclude that, even on changing the X and Y components of initial velocity, range remains constant.

Illustration Questions

For the given projectiles, which of the two will have the same range? (g=10 m/s2) For P1  ux = 200 m/s,  uy = 10 m/s For P2  ux = 200 m/s,  uy = 5 m/s For P3  ux = 100 m/s,  uy = 50 m/s For P4  ux = 10 m/s,  uy = 200 m/s

A P2  and P3

B P1  and P4

C P3  and P4

D P2  and P4

×

Range for particle 1

\(R=\dfrac {2u_{x} \cdot u_{y} } {g}\)

\(=\dfrac {2(200)\cdot (10)}{10}=400\,m\)

image

Range for particle 2

\(R=\dfrac {2u_{x} \cdot u_{y} } {g}\)

\(=\dfrac {2(200)\cdot (5)}{10}=200\,m\)

image

Range for particle 3

\(R=\dfrac {2u_{x} \cdot u_{y} } {g}\)

\(=\dfrac {2(100)\cdot (50)}{10}=1000\,m\)

 

image

Range for particle 4

\(R=\dfrac {2u_{x} \cdot u_{y} } {g}\)

\(=\dfrac {2(10)\cdot (200)}{10}=400\,m\)

image

P1 and P4 have the same range.

image

For the given projectiles, which of the two will have the same range? (g=10 m/s2) For P1  ux = 200 m/s,  uy = 10 m/s For P2  ux = 200 m/s,  uy = 5 m/s For P3  ux = 100 m/s,  uy = 50 m/s For P4  ux = 10 m/s,  uy = 200 m/s

A

P2  and P3

.

B

P1  and P4

C

P3  and P4

D

P2  and P4

Option B is Correct

Time of Flight when a Particle is Projected from Some Height at an Angle \(\theta\)

  • Consider a particle is projected from some height 'H' with a velocity 'u' at an angle '\(\theta\)' with the horizontal.
  • To calculate the time of flight, select Y-axis along the height 'H' and X-axis perpendicular to height 'H', as shown in figure.
  • Assume the given 2-D motion as a combination of two 1-D independent motions.
  • The moment particle lands at point B, the vertical component of its displacement becomes  – H.
  • For motion along Y-axis,

vertical component of initial velocity, \(u_y=u\,sin\theta\)

\(\Delta Y=-H\)

\(a_y=-g\)

By third law of motion;

\(\Delta Y=u_yt+\dfrac {1}{2}a_y(t)^2\)

\(\Rightarrow\) \(-H=(u\,sin\theta)\,t-\dfrac {1}{2}gt^2\)

Solve the equation for (t) to get the time of flight.

Illustration Questions

If a particle is projected from height H = 10 m at an angle \(\theta=\) 30° with initial velocity u = 10 m/s. Find the total time of flight. (g = 10 m/s2)

A 3 sec

B 2 sec

C 4 sec

D 1 sec

×

Height, H = 10 m

Velocity of particle, u = 10 m/s

Angle of projection, \(\theta\) = 30°

image

Vertical component of velocity,

\(u_y=u\,sin\theta\)

= 10 × sin 30°

= 5 m/s

image

Vertical component of acceleration \(a_y=-g\)

Height, \(\Delta Y=-10\;m\)

image

By third Law of motion,

\(\Delta Y=u_yt+\dfrac {1}{2}a_y(t)^2\)

 \(-10=5\,t-\dfrac {1}{2}(10)t^2\)

On solving for t,

t = 2 sec

image

If a particle is projected from height H = 10 m at an angle \(\theta=\) 30° with initial velocity u = 10 m/s. Find the total time of flight. (g = 10 m/s2)

image
A

3 sec

.

B

2 sec

C

4 sec

D

1 sec

Option B is Correct

Range of a Particle when Projected from some Height at an Angle \(\theta\)

  • Consider a particle is projected from some height 'H' at an angle \(\theta\) with the horizontal.
  • To calculate the range select Y-axis along the height 'H' and X-axis perpendicular to the height 'H' as shown in the figure.
  • Assume the given 2-D motion as a combination of two 1-D independent motions.
  • For motion along X-axis

X-component of initial velocity, \(u_x=u\,cos\,\theta\)

\(a_x=0\)

\(R=u\,cos\,\theta (t)\) ...(1)

  • For motion along Y-axis

Y-component of initial velocity, \(u_y=u\,sin\,\theta\)

\(\Delta Y=-H\)

\(a_y=-g\)

\(-H=(u\,sin\theta)t-\dfrac {1}{2}gt^2\)

  • Calculate t and put its value in equation (1), we get the value of range.

Illustration Questions

A particle is projected from height H = 10 m at an angle \(\theta=\)30° with initial velocity u = 10 m/s, find the range of particle.

A \(10\sqrt 3\) m

B \(20\sqrt 3\) m

C \(30\sqrt 3\) m

D \(10\) m

×

Calculation of Time of Flight

Height, H = 10 m

Velocity of particle, u = 10 m/s

Angle of Projection, \(\theta =\) 30°

Vertical component of velocity,

\(u_y=u\,sin\,\theta\)

= 10 sin 30°

= 5 m/s

\(a_y=-g\)

\(\Delta Y=-10\,m\)

image

By third law of motion,

\(\Delta Y=u_yt+\dfrac {1}{2}(a_y)t^2\)

\(-10=5t-\dfrac {1}{2}(10)t^2\)

t = 2 sec

Calculation of Range

Velocity of particle, u = 10 m/s

Angle of Projection, \(\theta =\) 30°

Horizontal component of velocity,

\(u_x=u\,cos\,\theta\)

= 10 cos 30°

= \(5\sqrt3\) m/s

image

Time taken by particle, 

t = 2 sec

Range (R) = \(u_x\,t\)

=\(10\sqrt3\) m

A particle is projected from height H = 10 m at an angle \(\theta=\)30° with initial velocity u = 10 m/s, find the range of particle.

image
A

\(10\sqrt 3\) m

.

B

\(20\sqrt 3\) m

C

\(30\sqrt 3\) m

D

\(10\) m

Option A is Correct

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