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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., $$a_y\,=\,-g$$.

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

Vertical component of velocity

$$v_y$$$$v\,sin\,\theta$$

$$v_y$$ = 5 × sin 37°

= 3 m/s

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

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

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

Y-component of initial velocity

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

$$=12 \,sin (75° – 15° )$$

$$= 12 ×$$ $$\dfrac {\sqrt 3}{2}$$

= $$6\sqrt 3$$ $$m/s$$

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

$$u_x=$$ $$6 \,m/s$$,  $$u_y=$$ $$6\sqrt 3$$ $$m/s$$

Option D 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.

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

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

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

Range for particle 2

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

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

Range for particle 3

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

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

Range for particle 4

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

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

P1 and P4 have the same range.

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.

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°

Vertical component of velocity,

$$u_y=u\,sin\theta$$

= 10 × sin 30°

= 5 m/s

Vertical component of acceleration $$a_y=-g$$

Height, $$\Delta Y=-10\;m$$

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

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

Option B 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

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

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°

Time of Flight

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

$$T=\dfrac {2(25)\;sin\,37°}{10}$$

= 3 sec

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

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°

$$R_{max}=\dfrac {u^2}{g}$$

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

Range of particle,

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

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

$$=\dfrac {625}{10}=62.5\;m$$

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  $$\Delta y=u_yt+\dfrac {1}{2}a_yt^2$$

$$0×t+\dfrac {1}{2}×(-g)(t^2)=-H$$

$$t=\sqrt {\dfrac {2H}{g}}$$

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

Height of building, H = 20 m

Time taken by particle is

$$t=\sqrt {\dfrac {2H}{g}}$$

$$t=\sqrt {\dfrac {2×(20)}{10}}$$

= 2 sec

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 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.

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

$$a_y=-g$$

$$\Delta Y=-10\,m$$

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

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.

A

$$10\sqrt 3$$ m

.

B

$$20\sqrt 3$$ m

C

$$30\sqrt 3$$ m

D

$$10$$ m

Option A 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-
• $$v^2=u^2+2as$$

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

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

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°

Maximum height attained by particle

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

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

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

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

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

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

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

Time of flight for particle 2

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

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

Time of flight for particle 3

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

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

Time of flight for particle 4

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

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

Time of flight for particle C is maximum.

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