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### Various Parameters Of Projectile

Practice to find maximum height of a projectile, vertical displacement of a projectile & velocity of a particle at a given instant. Learn formulas how to calculate of projectile motion equations.

# Position of a Particle at Given Instant

• A particle is projected with an initial velocity u, making an angle $$\theta$$ with the horizontal, as shown in figure.
• Point of projection of particle is considered as origin.

## Time of Flight

Time of flight of a particle projected along a horizontal plane

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

### Displacement along $$x\;and\ y$$ axis at time $$(t<T)$$

Here, t < T (Time of flight)

### Parameters along x-axis

• Initial velocity, $$u_x=u\,cos\,\theta$$
• Acceleration, $$a_x=0$$
• Displacement along x-axis at time t, x(t)

using  $$s=ut+\dfrac {1}{2}at^2$$

$$x(t)=u_xt+\dfrac {1}{2}a_xt^2$$

$$x(t)=u_xt$$

$$x(t)=(u\,cos\,\theta)\,t$$

### Parameters along y-axis

• Initial velocity, $$u_y=u\,sin\,\theta$$
• Acceleration, $$a_y=-g$$
• Displacement along y-axis at time 't', y(t) using

$$s=ut+\dfrac {1}{2}at^2$$

$$y(t)=u_yt+\dfrac {1}{2}a_yt^2$$

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

The acceleration along $$x\;and \;y$$ axis are shown in figure.

#### A particle is projected with velocity $$u=20\,m/s$$ , making an angle $$\theta=30°$$ with the horizontal. Determine position $$x$$ and $$y$$ at time $$t=1\,sec$$.  Assuming point of projection as origin. [ Take $$g=10\,m/s^2$$ ]

A $$x=10\sqrt 3\,m,\,\;y=5\,m$$

B $$x=10\,m,\,\;y=5\,\sqrt 3\,m$$

C $$x=5\,m,\,\;y=5\,m$$

D $$x=4\,m,\,\;y=3\,m$$

×

Given : Initial velocity, $$u=20\,m/s$$ , $$\theta=30°$$

Parameters along x-axis

$$u_x=u\,cos\,30°$$

$$=20.\dfrac {\sqrt 3} {2}$$

$$u_x=10\sqrt 3\;m/s$$

$$a_x=0$$

Parameters along y-axis

$$u_y=u\,sin\,30°$$

$$u_y$$ $$=20×\dfrac {1} {2}$$

$$u_y=10\;m/s$$

$$a_y=-g$$

Position of a particle at time t,

using  $$s(t)=ut+\dfrac {1}{2}at^2$$

$$x(t)=u_xt+\dfrac {1}{2}a_xt^2$$           (at  $$t=1\,sec$$ )

$$x(1)=10\sqrt 3×1+\dfrac {1}{2}×0×(1)^2$$

$$\Rightarrow x(1)=10\sqrt 3\;m$$

$$y(t)=u_yt+\dfrac {1}{2}a_yt^2$$

$$y(1)=10×1+\dfrac {1}{2}×(-g)×(1)^2$$

$$\Rightarrow y(1)=10-5\;$$

$$\Rightarrow y(1)=5\;m$$

### A particle is projected with velocity $$u=20\,m/s$$ , making an angle $$\theta=30°$$ with the horizontal. Determine position $$x$$ and $$y$$ at time $$t=1\,sec$$.  Assuming point of projection as origin. [ Take $$g=10\,m/s^2$$ ]

A

$$x=10\sqrt 3\,m,\,\;y=5\,m$$

.

B

$$x=10\,m,\,\;y=5\,\sqrt 3\,m$$

C

$$x=5\,m,\,\;y=5\,m$$

D

$$x=4\,m,\,\;y=3\,m$$

Option A is Correct

# Velocity of a Particle at Given Instant

• A particle is projected with an initial velocity u, making an angle $$\theta$$ with the horizontal, as shown in figure.
• Velocity of a particle at any time is v $$(t<T)$$, where T = time of flight.

## Parameters along x-axis

• Initial velocity, $$u_x=u\,\cos\theta$$
• Acceleration $$a_x=0$$
• Velocity along x-axis at time t,

using $$v=u+at$$

$$v_x=u_x+a_xt$$

$$v_x$$  $$=u_x+0.t$$

$$v_x=u_x$$

$$v_x=u\,cos\,\theta$$

## Parameters along y-axis

• Initial velocity, $$u_y=u\,\sin\theta$$
• Acceleration $$a_y=-g$$
• Velocity along y-axis at time t,

using $$v=u+at$$

$$v_y=u_y+a_yt$$

$$v_y=u_y-g\,t$$

$$v_y=u\,sin\,\theta-g\,t$$

### Velocity at time t

$$\vec v=v_x\hat i+v_y\hat j$$

$$\vec v=u_x\hat i+(u_y-g\,t)\,\hat j$$

$$|\vec v|=\sqrt {v_x^2+v_y^2}$$

#### A particle is projected with velocity $$u=20\,m/sec$$  making an angle $$\theta=30°$$ with the horizontal. Determine its velocity at time $$t=1\,sec$$. [ Take $$g=10\,m/s^2$$]

A $$5\, m/s$$

B $$4\, m/s$$

C $$10\sqrt 3\, m/s$$

D $$3\, m/s$$

×

Given:

Initial velocity $$u=20\, m/s$$$$\theta=30°$$

Parameters along x-axis

$$u_x=u\,cos\,30°$$

$$u_x$$ $$=20×\dfrac {\sqrt 3}{2}$$

$$u_x=10\sqrt 3\,m/s$$

$$a_x=0$$

Parameters along y-axis

$$u_y=u\,sin\,30°$$

$$=20×\dfrac {1}{2}$$

$$u_y=10\,m/s$$

$$a_y=-g$$

Velocity of particle at time t = 1 sec

using

$$v=u+at$$

$$v_x(t)=u_x+a_xt$$

$$v_x(1)=10\,\sqrt 3\,m/sec$$

$$v_y(t)=u_y+a_y.t$$

$$v_y(1)=10-10$$

$$v_y(1)=0\,m/sec$$

Velocity

$$\vec v=v_x\hat i+v_y\hat j$$

$$\vec v=10\sqrt 3\,\hat i$$

$$|\vec v|=10\sqrt 3\,m/sec$$

### A particle is projected with velocity $$u=20\,m/sec$$  making an angle $$\theta=30°$$ with the horizontal. Determine its velocity at time $$t=1\,sec$$. [ Take $$g=10\,m/s^2$$]

A

$$5\, m/s$$

.

B

$$4\, m/s$$

C

$$10\sqrt 3\, m/s$$

D

$$3\, m/s$$

Option C is Correct

# Kinetic Energy of a Particle at Highest Point

• A particle is projected with an initial velocity u, making an angle $$\theta$$ with the horizontal, as shown in figure.

### Parameters along x-axis

• Initial velocity  $$u_x=u\,cos\,\theta$$
• Acceleration  $$a_x=0$$

### Parameters along y-axis

• Initial velocity $$u_y=u\,sin\,\theta$$
• Acceleration $$a_y=-g$$

### At Highest point

• Velocity of a particle in y-direction $$v_y=0$$
• Velocity of a particle in x-direction $$v_x=u_x=u\,cos\,\theta$$
• Speed of particle at highest point $$=u\,cos\,\theta$$

As we know,

Kinetic Energy = $$\dfrac {1}{2}\,m\,v^2$$

Kinetic Energy(At highest point) = $$\dfrac {1}{2}\,m\;(u\,cos\,\theta)^2$$

#### A particle is projected with velocity u = 20 m/s, making an angle $$\theta$$ = 37° with the horizontal. If mass of the particle is 2 kg then determine its momentum at the highest point. [Take $$sin\,37°=\dfrac {3}{5}\,,\,cos\,37°=\dfrac {4}{5}$$]

A 16 N sec

B 20 N sec

C 15 N sec

D 32 N sec

×

Given:

Initial velocity, u = 20 m/s, $$\theta$$= 37°

Parameters along x-axis

$$u_x=u\,cos\,37°$$

$$=20×\dfrac {4}{5}$$

$$=16\, m/s$$

$$a_x=0$$

Parameters along y-axis

$$u_y=u\,sin\,37°$$

$$=20×\dfrac {3}{5}$$

$$=12\, m/s$$

$$a_y=-g$$

Momentum at highest point

$$|\vec P|=m\cdot v_x$$ (Because at highest point $$\vec {v_y}=0$$ and $$v_x=u_x$$)

$$=2×16$$

$$|\vec P|= 32\;N\,sec$$

### A particle is projected with velocity u = 20 m/s, making an angle $$\theta$$ = 37° with the horizontal. If mass of the particle is 2 kg then determine its momentum at the highest point. [Take $$sin\,37°=\dfrac {3}{5}\,,\,cos\,37°=\dfrac {4}{5}$$]

A

16 N sec

.

B

20 N sec

C

15 N sec

D

32 N sec

Option D is Correct

# Calculation of Vertical Displacement of a Projectile at a Given Instant

• A particle is projected with an initial velocity u, making an angle $$\theta$$ with the horizontal, as shown in figure.
• ## Time of Flight

Time of Flight  $$T=\dfrac {2\,u\,sin\,\theta}{g}$$

• To calculate displacement along y-axis or vertical displacement at  $$t<T$$.
• Initial velocity along y-direction $$v_y=u\,sin\;\theta$$
• Acceleration $$a_y=-g$$
• Displacement using $$s(t)=ut+\dfrac {1}{2}at^2$$
• $$y(t)=u\,sin\,\theta\,.t-\dfrac {1}{2}\,g\,t^2$$

#### A particle is projected with a velocity $$u = 50\; m/sec$$  making an angle $$\theta = 30°$$ with the horizontal. Calculate its vertical displacement at time $$t = 3\; sec$$.

A 10 m

B 20 m

C 15 m

D 30 m

×

Given:

Initial velocity $$u = 50\; m/sec$$ , $$\theta = 30°$$

Initial velocity along y-direction

$$v_y=u\;sin\;\theta$$

$$=50\;sin\;30°$$

$$=50×\dfrac{1}{2}$$

$$u_y=25\;m/sec$$

Acceleration, $$a_y$$ = – g

Vertical displacement at t = 3 sec

$$y(t)=u_yt-\dfrac {1}{2}\;gt^2$$

$$=25×3-\dfrac {1}{2}\;×10×(3)^2$$

$$=75-45$$

$$=30\;m$$

### A particle is projected with a velocity $$u = 50\; m/sec$$  making an angle $$\theta = 30°$$ with the horizontal. Calculate its vertical displacement at time $$t = 3\; sec$$.

A

10 m

.

B

20 m

C

15 m

D

30 m

Option D is Correct

# Calculation of Time at which the Projectile is at Height less than Maximum Height

## Maximum Height

• A particle is projected with initial velocity $$u$$, making an angle $$\theta$$ with the horizontal.

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

### Motion in y-direction (Vertical Displacement)

• Initial velocity in y-direction

$$u_y=u\;sin\;\theta$$

$$a_y=-g$$

### Vertical Displacement, at time t

using $$s=ut+\dfrac {1}{2}\,at^2$$       $$(\because\,\, h=s)$$

$$h=u\;sin\;\theta \cdot t-\dfrac {1}{2}\,gt^2$$

$$gt^2-2\;u\;sin\;\theta\cdot t+h=0$$

$$t=\dfrac {2\;u\;sin\;\theta\pm \sqrt {4\;u^2\;sin^2\;\theta-4gh}}{2g}$$

Here, two values of time indicating projectile is achieving same value of height h, for both the instants.

#### A particle is projected with a velocity $$u = 78\; m/s$$, making an angle $$\theta = 23°$$with the horizontal. Calculate the time when the particle is at height $$h = 40\; m$$.$$\Big [Take \,\,sin\;23°=\dfrac {5}{13}\Big]$$

A $$t=4\;sec,\;2\;sec$$

B $$t=5\;sec,\;8\;sec$$

C $$t=3\;sec,\;8\;sec$$

D $$t=8\;sec,\;4\;sec$$

×

Given:

Initial velocity $$u = 78\; m/s$$, $$\theta = 23°$$ and  $$h = 40\; m$$

$$u_y=u\sin\;\theta$$

$$=75×sin\,23°$$

$$=78 ×\dfrac{5}{13}$$

$$=30\;m/sec$$

$$a_y=-g$$

$$a_y=-10\;m/s^2$$

Use

$$s=ut+\dfrac {1}{2}\,at^2$$

$$\Rightarrow h=u_yt-\dfrac {1}{2}\;gt^2$$

$$\Rightarrow 40=30t-5t^2$$

$$\Rightarrow5t^2-30t+40=0$$

$$t=\dfrac {30\pm\sqrt {900-4×40×5}}{2×5}$$

$$t=\dfrac {30\pm10}{10}$$

$$t=\dfrac {40}{10},\;\dfrac {20}{10}$$

$$t=4\;sec,\;2\;sec$$

### A particle is projected with a velocity $$u = 78\; m/s$$, making an angle $$\theta = 23°$$with the horizontal. Calculate the time when the particle is at height $$h = 40\; m$$.$$\Big [Take \,\,sin\;23°=\dfrac {5}{13}\Big]$$

A

$$t=4\;sec,\;2\;sec$$

.

B

$$t=5\;sec,\;8\;sec$$

C

$$t=3\;sec,\;8\;sec$$

D

$$t=8\;sec,\;4\;sec$$

Option A is Correct

#### A particle is projected with velocity $$u = 20\; m/sec$$ making an angle of $$\theta = 37°$$ from a height of $$h = 20\; m$$, as shown in figure. Calculate the time when final velocity is perpendicular to the initial velocity.

A $$t=\dfrac {10}{3}\;sec$$

B $$t=\dfrac {20}{3}\;sec$$

C $$t=4\;sec$$

D $$t=2\;sec$$

×

Given,

Initial velocity $$u = 20\; m/sec$$$$\theta = 37°$$$$h = 20\; m$$

Initial velocity,

$$\vec u=u_x\hat i+u_y\hat j$$

$$=(u\;cos\;\theta)\hat i+(u\;sin\;\theta)\hat j$$

$$=(20\;cos\;37°)\hat i+(20\;sin\;37°)\hat j$$

$$=\left (20×\dfrac {4}{5}\right)\hat i+ \left (20×\dfrac {3}{5}\right)\hat j$$

$$\vec u=16\;\hat i+12\;\hat j$$

Final velocity at time t,

Use $$v=u+at$$

$$\vec v=\vec u+\vec {a_y}\;t$$

$$\vec v=(u_x\hat i+u_y\hat j)-g\;t\;\hat j$$

$$\vec v=16\;\hat i+(12-10t)\;\hat j$$

For $$\vec v \perp\vec u$$

$$\\\;\vec u.\vec v=0$$

$$(16\;\hat i+12\;\hat j)\cdot [16\;\hat i+(12-10t)\;\hat j]=0$$

$$(16)^2+(12)^2-120\;t=0$$

$$256+144-120\,t=0$$

$$400=120\,t$$

$$t=\dfrac {10}{3}\;sec$$

### A particle is projected with velocity $$u = 20\; m/sec$$ making an angle of $$\theta = 37°$$ from a height of $$h = 20\; m$$, as shown in figure. Calculate the time when final velocity is perpendicular to the initial velocity.

A

$$t=\dfrac {10}{3}\;sec$$

.

B

$$t=\dfrac {20}{3}\;sec$$

C

$$t=4\;sec$$

D

$$t=2\;sec$$

Option A is Correct