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### Wind And Projection On Inclined Plane

Practice to calculation of time of flight, maximum height, range, projection up an inclined plane, and maximum range along an Incline.

# Calculation of Time of Flight

• Consider a particle, projected with an initial velocity $$u$$, making an angle $$\theta$$ with the horizontal direction.

The particle has two accelerations :

1. Acceleration due to gravity '$$g$$' in downward direction.
2. Horizontal acceleration '$$a$$'.
• Point of projection of the particle is considered as origin.

• As the effect of any vector in perpendicular direction is zero, so all vectors are resolved along $$x$$ and $$y$$ directions which are mutually perpendicular.
• Time is a scalar quantity, so its components cannot be taken along $$x-y$$  axis.
• It is the only parameter which relates all the other parameters.

Along $$x$$ direction :

• Initial velocity, $$u_x=u\,cos\,\theta$$
• Horizontal acceleration , $$a_x=a$$

Along $$y$$ direction :

• Initial velocity, $$u_y=u\,sin\,\theta$$
• Acceleration,  $$a_y=-g$$
• The time in which the projected particle reaches to the plane of projection again, is known as the time of flight.

Using,

$$s_y=u_yt+\dfrac{1}{2}\,a_yt^2$$

At time of flight $$T$$

$$s_y=0,\;t=T$$

$$0=u_y\,T+\dfrac{1}{2}a_y\,T^2$$

$$0=u\,sin\,\theta.T+\dfrac{1}{2}(-g)\,T^2$$

$$T=\dfrac{2usin\,\theta}{g}$$

$$T=\left|\dfrac{2\,u_y}{g}\right|$$

Note : The equation implies that the time of flight is independent of the acceleration along horizontal direction.

#### A particle is projected with an initial velocity $$u=30\;m/sec$$ at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing which gives a constant acceleration  $$a=5\;m/s^2$$ to the particle in horizontal direction, as shown in the figure. Calculate the time of flight of the particle. [Take $$g=10\;m/s^2$$]

A $$4\,sec$$

B $$5\,sec$$

C $$3\,sec$$

D $$6\,sec$$

×

Taking components along $$x \;and \;y$$ axis

Along $$x-$$ axis

$$u_x=u\,cos\,30°$$

$$u_x=30\,cos\,30°$$

$$=\dfrac{30\sqrt3}{2}\,m/s$$

$$a_x=5\;m/s^2$$

Along $$y-$$ axis

$$u_y=u\,sin\,30°$$

$$u_y=30\,sin\,30°$$

$$=30×\dfrac{1}{2}$$

$$=15\;m/s$$

$$a_y=-g$$

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

Time of flight,

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

where, $$u\,sin\theta=15$$

$$T=\dfrac{2×15}{10}$$

$$T=3\,sec$$

### A particle is projected with an initial velocity $$u=30\;m/sec$$ at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing which gives a constant acceleration  $$a=5\;m/s^2$$ to the particle in horizontal direction, as shown in the figure. Calculate the time of flight of the particle. [Take $$g=10\;m/s^2$$]

A

$$4\,sec$$

.

B

$$5\,sec$$

C

$$3\,sec$$

D

$$6\,sec$$

Option C is Correct

# Maximum Height

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

The particle has two accelerations :

1. Acceleration due to gravity $$'g'$$ in downward direction.
2. Horizontal acceleration $$'a'$$.
• Point of projection of the particle is considered to be origin.

• As the effect of any vector in perpendicular direction is zero, so all vectors are resolved along $$x$$ and $$y$$ directions which are mutually perpendicular.
• Time is a scalar quantity, so its components cannot be taken along $$x-y$$ axis.
• It is the only parameter, which relates all the other parameters.

Along $$x$$ direction

• Initial velocity, $$u_x=u\,cos\,\theta$$
• Horizontal acceleration , $$a_x=a$$

Along $$y$$ direction

• Initial velocity, $$u_y=u\,sin\,\theta$$
• Acceleration, $$a_y=-g$$
• When particle reaches the maximum height, $$v_y$$ becomes zero.

Using $$v_y^2=u^2_y+2\,a_y\;s_y$$

when $$v_y=0,\;s_y=H_{max}$$

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

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

$$H_{max}=\left|\dfrac{u^2_y}{2\,a_y}\right|$$

#### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing which gives a constant acceleration $$a=5\,m/s^2$$ to the particle in horizontal direction, as shown in the figure. Calculate maximum height of the projectile. [Take $$g=10\,m/s^2$$]

A $$11.25\,m$$

B $$400\,m$$

C $$13\,m$$

D $$15\,m$$

×

Component along $$x$$ axis

$$u_x=30\,cos\,30°$$

$$=\dfrac{30\sqrt3}{2}=15\sqrt3\,m/s$$

$$a_x=5\,m/s^2$$

Component along $$y$$ axis

$$u_y=30\,sin\,30°$$

$$=\dfrac{30×1}{2}=15\,m/s$$

$$a_y=-g\,=-10\,m/s^2$$

Maximum Height

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

$$=\Bigg|\dfrac{u^2_y}{2g}\Bigg|$$

$$=\dfrac{225}{20}=11.25\,m$$

### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing which gives a constant acceleration $$a=5\,m/s^2$$ to the particle in horizontal direction, as shown in the figure. Calculate maximum height of the projectile. [Take $$g=10\,m/s^2$$]

A

$$11.25\,m$$

.

B

$$400\,m$$

C

$$13\,m$$

D

$$15\,m$$

Option A is Correct

# Projection up an Inclined Plane

• Suppose an inclined plane is making an angle $$'\alpha'$$ with the horizontal direction.
• A particle is projected with speed $$u$$ at angle $$\theta$$ on the inclined plane, as shown in figure.
• Take $$x-$$ axis along inclined plane and $$y-$$ axis perpendicular to it.
• The point of projection is considered as origin.

• Resolve the components along $$x$$ and $$y$$ axis.

Along $$x-$$ axis

$$u_x=u\,cos\,\theta$$

$$a_x=-g\,sin\,\alpha$$

Along $$y-$$ axis

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

$$a_y=-g\,cos\,\alpha$$

### Time of flight

• The time in which the projected particle strikes the inclined plane is called time of flight.

using

$$s_y=u_yt+\dfrac{1}{2}\,a_yt^2$$

Let the time of flight be given by $$T$$.

At $$t=T$$, the displacement in $$y$$ direction will be zero.

$$0=u_yT+\dfrac{1}{2}\,a_yT^2$$

$$0=u\,sin\,\theta\;T+\dfrac{1}{2}\,(-g\,cos\,\alpha)\,T^2$$

$$T=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}=\left|\dfrac{2\,u_y}{a_y}\right|$$

### Maximum height

• Maximum displacement perpendicular to the inclined plane is known as maximum height.
• At maximum height above the inclined plane, velocity in $$y$$ direction is zero.

Let maximum height be given by $$H_{max}$$.

using,   $$v^2_y=u^2_y+2a_ys_y$$

$$0=u^2sin^2\theta-2g\,cos\,\alpha\;H_{max}$$

$$H_{max}=\dfrac{u^2sin^2\theta}{2g\,cos\,\alpha}=\dfrac{u^2_y}{2|a_y|}$$

#### A particle is projected up an inclined plane with an initial velocity $$u=40\,m/s$$, at an angle $$\theta=30°$$. Plane is inclined at an angle $$\alpha=37°$$ with the horizontal. Determine the time of flight and the maximum range of the particle. $$\left[\text{Take }:sin\,37°=\dfrac{3}{5},\;g=10\,m/s^2\right]$$

A $$T=5\,sec,\;H=25\,m$$

B $$T=8\,sec,\;H=5\,m$$

C $$T=7\,sec,\;H=10\,m$$

D $$T=9\,sec,\;H=12\,m$$

×

Initial velocity in $$y$$ direction

$$u\,sin\,\theta=u\,sin\,30°$$

$$=40\,sin\,30°$$

$$=20\,m/s=u_y$$

Acceleration in $$y$$ direction

$$g\,cos\,\alpha=10\,cos\,37°$$

$$=10×\dfrac{4}{5}$$

$$=8\,m/s^2=a_y$$

Time of flight for projection up an inclined plane,

$$T=\left|\dfrac{2\,u_y}{a_y}\right|=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}$$

[where $$u_y=u\,sin\,\theta$$]

$$T=\dfrac{2×20}{8}$$

$$T=5\,sec$$

Maximum height, $$H_{max}=\dfrac{(u\,sin\,\theta)^2}{2g\,cos\,\alpha}=\dfrac{u^2_y}{2|a_y|}$$

[where $$u_y=u\,sin\,\theta$$]

$$=\dfrac{(20)^2}{2×8}$$

$$=\dfrac{400}{16}=25\,m$$

### A particle is projected up an inclined plane with an initial velocity $$u=40\,m/s$$, at an angle $$\theta=30°$$. Plane is inclined at an angle $$\alpha=37°$$ with the horizontal. Determine the time of flight and the maximum range of the particle. $$\left[\text{Take }:sin\,37°=\dfrac{3}{5},\;g=10\,m/s^2\right]$$

A

$$T=5\,sec,\;H=25\,m$$

.

B

$$T=8\,sec,\;H=5\,m$$

C

$$T=7\,sec,\;H=10\,m$$

D

$$T=9\,sec,\;H=12\,m$$

Option A is Correct

# Projection Down an Inclined Plane

• Suppose an inclined plane is making an angle $$'\alpha'$$ with horizontal.
• A particle is projected with speed $$u$$ down the inclined plane.
• The particle is making an angle $$\theta$$ with the slope of the inclined plane, as shown in figure.

• Assumptions
1.  The point of projection is considered as origin.
2.  Consider $$x-$$ axis in downward direction along the slope of inclined plane and take $$y-$$ axis perpendicular to it.
• Components along $$x-$$ axis

Initial velocity, $$u_x=u\,cos\,\theta$$

Acceleration, $$a_x=g\,sin\,\alpha$$

• Components along $$y-$$ axis

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

Acceleration, $$a_y=-g\,cos\,\alpha$$

### Time of flight

• The time taken by the particle to return to the inclined plane is known as time of flight.

using

$$s_y=u_yt+\dfrac{1}{2}\,a_yt^2$$

At $$t=T$$, the displacement in $$y$$ direction will be zero.

$$0=u_yT+\dfrac{1}{2}\,a_yT^2$$

$$0=u\,sin\,\theta\;T+\dfrac{1}{2}(-g\,cos\,\alpha)\,T^2$$

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

$$T=\left|\dfrac{2\,u_y}{a_y}\right|$$

### Maximum height

• Maximum displacement perpendicular to the inclined plane is known as maximum height.

Let maximum height be $$H_{max}$$.

using,  $$v_y^2=u_y^2+2a_ys_y$$

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

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

$$H_{max}=\dfrac{u_y^2}{2|a_y|}$$

#### A particle is projected down an inclined plane with velocity $$u=30\,m/s$$ making an angle $$\theta=30°$$ with inclined plane. The plane is inclined at an angle $$\alpha=60°$$ with the horizontal. Determine the time of flight and maximum height of the particle.  $$[\text{Take }:g=10\,m/s^2]$$

A $$6\,sec,\;22.5\,m$$

B $$7\,sec,\;30\,m$$

C $$8\,sec,\;25\,m$$

D $$9\,sec,\;20\,m$$

×

For projection down an inclined plane,

Time of flight $$=\left|\dfrac{2u_y}{a_y}\right|$$ and

Maximum height $$=\dfrac{u_y^2}{2\,|a_y|}$$

here, $$u_y=u\,sin\,30°$$

$$=30\,sin\,30°=15\,m/s$$ and

$$|a_y|=10\,cos\,60°=5\,m/s^2$$

Time of flight,  $$T=\left|\dfrac{2\,u_y}{a_y}\right|$$

$$=\dfrac{2×15}{5}$$

$$=6\,sec$$

Maximum height,

$$(H_{max})=\dfrac{u_y^2}{2\,|a_y|}$$

$$H_{max}=\dfrac{(15)^2}{2×5}$$

$$=22.5\,m$$

### A particle is projected down an inclined plane with velocity $$u=30\,m/s$$ making an angle $$\theta=30°$$ with inclined plane. The plane is inclined at an angle $$\alpha=60°$$ with the horizontal. Determine the time of flight and maximum height of the particle.  $$[\text{Take }:g=10\,m/s^2]$$

A

$$6\,sec,\;22.5\,m$$

.

B

$$7\,sec,\;30\,m$$

C

$$8\,sec,\;25\,m$$

D

$$9\,sec,\;20\,m$$

Option A is Correct

# Calculation of Range

• Consider a particle, projected with an initial velocity $$u$$, making an angle $$\theta$$ with the horizontal direction.

The particle has two accelerations

1. Acceleration due to gravity $$'g'$$ in downward direction.
2. Horizontal acceleration $$'a'$$.
• Point of projection of the particle is considered as origin.

• As the effect of any vector in perpendicular direction is zero, so all vectors are resolved along $$x$$ and $$y$$ directions which are mutually perpendicular.
• Time is a scalar quantity, so its components can not be taken along $$x$$ and $$y$$ axis.
• It is the only parameter which relates all the other parameters.

Along $$x$$ direction

• Initial velocity $$u_x=u\,cos\,\theta$$
• Horizontal acceleration $$a_x=a$$

Along $$y$$ direction

• Initial velocity $$u_y=u\,sin\,\theta$$
• Acceleration $$a_y=-g$$
• Total displacement covered by the particle along $$x-$$ axis is known as the range of the projectile.

Using,

$$s_x=u_xt+\dfrac{1}{2}a_xt^2$$

where,  $$s_x=$$ Total displacement along $$x-$$ axis $$=R$$

$$t=\;T=$$ time of flight

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

$$R=u\,cos\,\theta\,\left(\dfrac{2u\,sin\,\theta}{g}\right)+\dfrac{1}{2}a\,\left(\dfrac{2u\,sin\,\theta}{g}\right)^2$$

$$R=\dfrac{u^2\,sin2\,\theta}{g}+\dfrac{2au^2}{g^2}\,sin^2\theta$$

#### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal plane. Simultaneously strong wind starts blowing which gives a constant acceleration $$a=5\,m/s^2$$ to the particle in horizontal direction. Determine the horizontal range of the particle. [Take $$g=10\,m/s^2$$]

A $$92\,m$$

B $$90\,m$$

C $$99\,m$$

D $$100.44\,m$$

×

Along $$x$$ axis

$$u_x=30\,cos\,30°$$

$$=\dfrac{30×\sqrt3}{2}=15\sqrt3\,m/s$$

$$a_x=5\,m/s^2$$

Along $$y$$ axis

$$u_y=30\,sin\,30°$$

$$=30×\dfrac{1}{2}=15\,m/s$$

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

Calculation of time of flight

$$T=\left|\dfrac{2u_y}{g}\right|$$

$$=\dfrac{2×15}{10}$$

$$=3\,sec$$

Horizontal range

$$R=u_x\,T+\dfrac{1}{2}\,a_x\,T^2$$

$$R=15\sqrt3×3+\dfrac{1}{2}×5×3^2$$

$$=45\sqrt3+\dfrac{45}{2}$$

$$R=45\,(2.23)=100.44\,m$$

### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal plane. Simultaneously strong wind starts blowing which gives a constant acceleration $$a=5\,m/s^2$$ to the particle in horizontal direction. Determine the horizontal range of the particle. [Take $$g=10\,m/s^2$$]

A

$$92\,m$$

.

B

$$90\,m$$

C

$$99\,m$$

D

$$100.44\,m$$

Option D is Correct

# Condition for Returning of Projectile at the Point of Projection (Range = 0)

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

The particle has two accelerations

1. Acceleration due to gravity $$'g'$$ in downward direction.
2. Horizontal acceleration $$'a_x\,'$$.
• Point of projection of the particle is considered to be origin as shown in the figure.

• As the effect of any vector in perpendicular direction is zero, so all vectors are resolved along $$x$$ and $$y$$ directions which are mutually perpendicular.
• Time is a scalar quantity, so its components can not be taken along $$x$$ and $$y$$ axis.
• It is the only parameter, which relates all the other parameters.

Along $$x-$$ direction

• Initial velocity,  $$u_x=u\,cos\,\theta$$
• Horizontal acceleration $$=a_x$$

Along $$y-$$ direction

• Initial velocity, $$u_y=u\,sin\,\theta$$
• Acceleration, $$a_y=-g$$
• Range of the projectile,

$$R=u_x\,T+\dfrac{1}{2}a_x\,T^2$$

where, $$T=$$ Time of flight

put $$T=\dfrac{2u\,sin\,\theta}{g}$$

$$R=u\,cos\,\theta\;\dfrac{2u\,sin\,\theta}{g}+\dfrac{1}{2}a_x\left(\dfrac{2u\,sin\,\theta}{g}\right)^2$$

$$R=\dfrac{u^2sin2\,\theta}{g}+2\,u^2sin^2\theta\;\dfrac{a_x}{g^2}$$

for range to be zero.

$$\Rightarrow\;\dfrac{u^2sin2\theta}{g}+2\,u^2sin^2\theta\;\dfrac{a_x}{g^2}=0$$

$$\Rightarrow\dfrac{u^2sin2\theta}{g}=-2u^2sin^2\theta\;\dfrac{a_x}{g^2}$$

$$\Rightarrow\;tan\,\theta=\dfrac{-g}{a_x}$$

This implies that acceleration $$'a_x\,'$$ should be along negative $$x$$ axis.

• To return at the point of projection, the net acceleration of the projectile should be in the opposite direction of the velocity.

#### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing, which gives a constant acceleration $$'a'$$ to the particle in horizontal direction as shown in the figure. What should be the value of acceleration $$'a'$$ provided by wind so that the particle could return to its point of projection? $$[g=10\,m/s^2]$$

A $$\dfrac{\sqrt3}{10}\ m/s^2$$

B $$\dfrac{10}{\sqrt3}\ m/s^2$$

C $$-10\sqrt3\ m/s^2$$

D $$10\ m/s^2$$

×

For returning at its point of projection i.e., range $$R=0$$

$$a_{net}$$ should make an angle $$\theta=30°$$ with the horizontal.

$$tan\,30°=\dfrac{-g}{a}$$

$$\dfrac{1}{\sqrt3}=\dfrac{-10}{a}$$

$$a=-10\sqrt3\ m/s^2$$

### A particle is projected with an initial velocity $$u=30\,m/s$$, at an angle $$\theta=30°$$ above the horizontal. Simultaneously strong wind starts blowing, which gives a constant acceleration $$'a'$$ to the particle in horizontal direction as shown in the figure. What should be the value of acceleration $$'a'$$ provided by wind so that the particle could return to its point of projection? $$[g=10\,m/s^2]$$

A

$$\dfrac{\sqrt3}{10}\ m/s^2$$

.

B

$$\dfrac{10}{\sqrt3}\ m/s^2$$

C

$$-10\sqrt3\ m/s^2$$

D

$$10\ m/s^2$$

Option C is Correct

# Maximum Range along an Incline

• Range of a particle is the distance between point of projection and the point where the particle hits the inclined plane.
• To calculate maximum range along an inclined plane, consider two situations :
1. When the particle is projected up the inclined plane.
2. When the particle is projected down the inclined plane.

Case 1

First we take the particle projected up an inclined plane.

$$x-$$ direction

Initial velocity , $$u_x=u\,cos\,\theta$$

Acceleration , $$a_x=-g\,sin\,\alpha$$

$$y-$$ direction

Initial velocity , $$u_y=u\;sin\; \theta$$

Acceleration , $$a_y=-g\,cos\,\alpha$$

Here, range = OA

Motion along $$y-$$ axis

When particle reaches at $$A$$, displacement along $$y-$$ axis becomes zero.

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

$$a_y=-g\,cos\,\alpha$$

$$\Delta y=0$$

$$\Delta y=u_yt+\dfrac{1}{2}a_y\,t^2$$

$$0=u\,sin\,\theta\;t+\dfrac{1}{2}\,(-g\,cos\,\alpha)\,t^2$$

$$0=u\,sin\,\theta\;t-\dfrac{1}{2}\,g\,cos\,\alpha\,t^2$$

$$t=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}$$

= time of flight $$(T)$$

Calculation of Range $$(R)$$

$$u_x=u\,cos\,\theta$$

$$a_x=-g\,sin\,\alpha$$

$$\Delta x=u_x\;t+\dfrac{1}{2}\,a_x\;t^2$$

At  $$t=T$$

$$\Delta x=R=u\,cos\,\theta\;T-\dfrac{1}{2}\,g\,sin\,\alpha\,T^2$$

$$R=u\,cos\,\theta\;\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)-\dfrac{1}{2}\,g\,sin\,\alpha\,\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)^2$$

$$R=\dfrac{2u^2sin\,\theta}{g\,cos^2\,\alpha}\,cos\,(\theta+\alpha)$$

Case 2

Now the particle is projected down the inclined plane.

$$x-$$ direction

Initial velocity , $$u_x=u\,cos\,\theta$$

Acceleration , $$a_x=g\,sin\,\alpha$$

$$y-$$ direction

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

Acceleration , $$a_y=-g\,cos\,\alpha$$

Now range = OA

Motion along $$y-$$ axis

When particle reaches at $$A$$, displacement along $$y-$$ axis becomes zero.

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

$$a_y=-g\,cos\,\alpha$$

$$\Delta y=0$$

$$\Delta y=u_y\;t+\dfrac{1}{2}a_y\,t^2$$

$$0=u\,sin\,\theta. t-\dfrac{1}{2}\,(g\,cos\,\alpha)\,t^2$$

$$t=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}$$

= time of flight $$(T)$$

Calculation of Range $$(R)$$

$$u_x=u\,cos\,\theta$$

$$a_x=g\,sin\,\alpha$$

$$\Delta x=u_x.t+\dfrac{1}{2}\,a_x\;t^2$$

At $$t=T$$

$$\Delta x=R=u\,cos\,\theta.\;T+\dfrac{1}{2}\,g\,sin\,\alpha\,T^2$$

$$R=u\,cos\,\theta\;\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)+\dfrac{1}{2}\,g\,sin\,\alpha\,\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)^2$$

$$R=\dfrac{2u^2sin\,\theta}{g\,cos^2\,\alpha}\,cos\,(\alpha-\theta)$$

#### A particle is projected up an inclined plane with an initial velocity $$u=40\,m/s$$ making an angle $$\theta=37°$$ with the inclined plane. The plane is inclined at an angle $$\alpha=37°$$ with the horizontal. Determine the range of the particle. $$\left[\text{Take }:sin\,37°=\dfrac{3}{5},\,\,\,cos\,37°=\dfrac{4}{5},\;g=10\;m/s^2\;\right]$$

A $$6\,sec,\;84\,m$$

B $$5\,sec,\;225\,m$$

C $$8\,sec,\;400\,m$$

D $$9\,sec,\;100\,m$$

×

The displacement along $$x-$$ direction during the time of flight = Range of projectile

Time of flight $$=\left|\dfrac{2\,u_y}{a_y}\right|$$

$$=\dfrac{2×u\,sin\,37°}{g\,cos\,37°}$$

$$=\dfrac{2×40×\dfrac{3}{5}}{10×\dfrac{4}{5}}$$

$$=\dfrac{48}{8}$$

$$T=6\,sec$$

Range = displacement along $$x$$ direction in $$6\,sec$$

using,

$$s_x=u_x\,T+\dfrac{1}{2}\,a_x\,T^2$$

$$a_x=-g\,sin\,\alpha$$

$$=-10×\dfrac{3}{5}=-6$$

$$\therefore \,s_x=40×\dfrac{4}{5}×6+\dfrac{1}{2}×(-6)×(6)^2$$

$$=192-108$$

$$=84\,m$$

### A particle is projected up an inclined plane with an initial velocity $$u=40\,m/s$$ making an angle $$\theta=37°$$ with the inclined plane. The plane is inclined at an angle $$\alpha=37°$$ with the horizontal. Determine the range of the particle. $$\left[\text{Take }:sin\,37°=\dfrac{3}{5},\,\,\,cos\,37°=\dfrac{4}{5},\;g=10\;m/s^2\;\right]$$

A

$$6\,sec,\;84\,m$$

.

B

$$5\,sec,\;225\,m$$

C

$$8\,sec,\;400\,m$$

D

$$9\,sec,\;100\,m$$

Option A is Correct

#### An archer standing on a $$30°$$ slope, shoots an arrow at $$30°$$above the horizontal, as shown in the figure. How far down the slope does the arrow hit, if it is projected with a velocity of $$50\,m/s$$? [Assuming the height of the man as negligible]  Take $$g=10\,m/s^2$$

A $$10\,sec,\;500\,m$$

B $$12\,sec,\;400\,m$$

C $$15\,sec,\;300\,m$$

D $$17\,sec,\;600\,m$$

×

When arrow hits the inclined plane, its displacement along $$y-$$ direction is zero, i.e., $$t=T$$ (time of flight)

$$\Rightarrow\;s_y=0$$

$$T=\left|\dfrac{2\,u_y}{a_y}\right|=\dfrac{2×50\,sin\,60°}{10\,cos\,30°}$$

$$=\dfrac{2×25\sqrt3}{5\sqrt3}$$

$$=10\,sec$$

Range = displacement along $$x$$ direction in time $$t=10\,sec$$

using,

$$s_x=u_xt+\dfrac{1}{2}a_xt^2$$

$$=50\,cos\,60°×10+\dfrac{1}{2}×g\,sin\,30°×(10)^2$$

$$=250+250$$

$$=500\,m$$

### An archer standing on a $$30°$$ slope, shoots an arrow at $$30°$$above the horizontal, as shown in the figure. How far down the slope does the arrow hit, if it is projected with a velocity of $$50\,m/s$$? [Assuming the height of the man as negligible]  Take $$g=10\,m/s^2$$

A

$$10\,sec,\;500\,m$$

.

B

$$12\,sec,\;400\,m$$

C

$$15\,sec,\;300\,m$$

D

$$17\,sec,\;600\,m$$

Option A is Correct