• 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.

• 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 \(v_y=0,\;s_y=H_{max}\)

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

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

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

• 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,   \(v^2_y=u^2_y+2a_ys_y\)

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

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

• 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,  \(v_y^2=u_y^2+2a_ys_y\)

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

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

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

• 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

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

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

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

• 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.

• 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 \(t=T\)

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

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

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