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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.

Illustration Questions

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

\(u_y=u\,sin\,30°\)

\(u_y=30\,sin\,30°\)

    \(=30×\dfrac{1}{2}\)

    \(=15\;m/s\)

\(a_y=-g\)

    \(=-10\;m/s^2\)

image

Time of flight,

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

 where, \(u\,sin\theta=15\) 

\(T=\dfrac{2×15}{10}\)

\(T=3\,sec\)

image

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 \(g=10\;m/s^2\)]

image
A

\(4\,sec\)

.

B

\(5\,sec\)

C

\(3\,sec\)

D

\(6\,sec\)

Option C 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

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

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

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

Illustration Questions

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

\(u_y=30\,sin\,30°\)

\(=30×\dfrac{1}{2}=15\,m/s\)

\(a_y=-10\,m/s^2\)

image

Calculation of time of flight

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

\(=\dfrac{2×15}{10}\)

\(=3\,sec\)

image

Horizontal range

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

\(R=15\sqrt3×3+\dfrac{1}{2}×5×3^2\)

\(=45\sqrt3+\dfrac{45}{2}\)

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

image

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 \(g=10\,m/s^2\)]

image
A

\(92\,m\)

.

B

\(90\,m\)

C

\(99\,m\)

D

\(100.44\,m\)

Option D 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 \(v_y=0,\;s_y=H_{max}\)

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

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

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

Illustration Questions

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

\(u_y=30\,sin\,30°\)

\(=\dfrac{30×1}{2}=15\,m/s\)

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

image

Maximum Height

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

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

\(=\dfrac{225}{20}=11.25\,m\)

image

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 \(g=10\,m/s^2\)]

image
A

\(11.25\,m\)

.

B

\(400\,m\)

C

\(13\,m\)

D

\(15\,m\)

Option A 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.

Illustration Questions

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.

\(tan\,30°=\dfrac{-g}{a}\)

\(\dfrac{1}{\sqrt3}=\dfrac{-10}{a}\)

\(a=-10\sqrt3\ m/s^2\)

image

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? \([g=10\,m/s^2]\)

image
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

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,   \(v^2_y=u^2_y+2a_ys_y\)

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

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

Illustration Questions

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

\(g\,cos\,\alpha=10\,cos\,37°\)

\(=10×\dfrac{4}{5}\)

\(=8\,m/s^2=a_y\)

image

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

\(T=\dfrac{2×20}{8}\)

\(T=5\,sec\)

image

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

[where \(u_y=u\,sin\,\theta\)]

\(=\dfrac{(20)^2}{2×8}\)

\(=\dfrac{400}{16}=25\,m\)

image

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. \(\left[\text{Take }:sin\,37°=\dfrac{3}{5},\;g=10\,m/s^2\right]\)

image
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,  \(v_y^2=u_y^2+2a_ys_y\)

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

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

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

Illustration Questions

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 

\(|a_y|=10\,cos\,60°=5\,m/s^2\)

image

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

\(=\dfrac{2×15}{5}\)

\(=6\,sec\)

image

Maximum height,

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

\(H_{max}=\dfrac{(15)^2}{2×5}\)

\(=22.5\,m\)

image

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.  \([\text{Take }:g=10\,m/s^2]\)

image
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

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 \(t=T\)

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

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

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

Illustration Questions

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

image

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

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

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

\(=\dfrac{48}{8}\)

\(T=6\,sec\)

image

Range = displacement along \(x\) direction in \(6\,sec\)

using,

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

\(a_x=-g\,sin\,\alpha\)

\(=-10×\dfrac{3}{5}=-6\)

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

\(=192-108\)

\(=84\,m\)

image

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. \(\left[\text{Take }:sin\,37°=\dfrac{3}{5},\,\,\,cos\,37°=\dfrac{4}{5},\;g=10\;m/s^2\;\right]\)

image
A

\(6\,sec,\;84\,m\)

.

B

\(5\,sec,\;225\,m\)

C

\(8\,sec,\;400\,m\)

D

\(9\,sec,\;100\,m\)

Option A is Correct

Illustration Questions

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 \(17\,sec,\;600\,m\)

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

\(\Rightarrow\;s_y=0\)

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

\(=\dfrac{2×25\sqrt3}{5\sqrt3}\)

\(=10\,sec\)

image

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

using,

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

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

\(=250+250\)

\(=500\,m\)

image

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 \(g=10\,m/s^2\)

image
A

\(10\,sec,\;500\,m\)

.

B

\(12\,sec,\;400\,m\)

C

\(15\,sec,\;300\,m\)

D

\(17\,sec,\;600\,m\)

Option A is Correct

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