It is the maximum vertical distance from the point of projection that a projectile can reach.

Using the third equation of motion for vertical motion ($v^2 = u^2 + 2as$), at the maximum height, the vertical component of velocity becomes zero ($v_y = 0$). The initial vertical velocity is $u_y = u \sin \theta$, and acceleration is $a_y = -g$:

$0^2 = (u \sin \theta)^2 + 2(-g)H$
$0 = u^2 \sin^2 \theta – 2gH$
$2gH = u^2 \sin^2 \theta$
$H = \frac{u^2 \sin^2 \theta}{2g}$

Important Notes on Maximum Height:

• Maximum height can also be expressed as:

  $H = \frac{u_y^2}{2g}$ (where $u_y$ is the vertical component of initial velocity).

• The absolute maximum height ($H_{max}$) is achieved when $\sin^2 \theta = 1$, which implies $\sin \theta = 1$ or $\theta = 90^\circ$.

  $H_{max} = \frac{u^2}{2g}$ (when $\theta = 90^\circ$)

  This means for maximum height, the body should be projected vertically upward. In this case, it falls back to the point of projection after reaching the maximum height.
• For complementary angles of projection $\theta$ and $90^\circ – \theta$:

  Ratio of maximum height: $\frac{H_1}{H_2} = \frac{u^2 \sin^2 \theta / 2g}{u^2 \sin^2(90^\circ – \theta) / 2g}$
  $\frac{H_1}{H_2} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$

If a particle passes through two points situated at the same height ‘$y$’ at times $t_1$ and $t_2$, then we can establish relationships between these times and the height.

Using the vertical motion equation $y = (u \sin \theta) t – \frac{1}{2} g t^2$:

$y = (u \sin \theta) t_1 – \frac{1}{2} g t_1^2 \quad \cdots (i)$
$y = (u \sin \theta) t_2 – \frac{1}{2} g t_2^2 \quad \cdots (ii)$

Comparing equation (i) and (ii):

$(u \sin \theta) t_1 – \frac{1}{2} g t_1^2 = (u \sin \theta) t_2 – \frac{1}{2} g t_2^2$
$(u \sin \theta) (t_1 – t_2) = \frac{1}{2} g (t_1^2 – t_2^2)$
$(u \sin \theta) (t_1 – t_2) = \frac{1}{2} g (t_1 – t_2)(t_1 + t_2)$

Since $t_1 \ne t_2$, we can divide by $(t_1 – t_2)$:

$u \sin \theta = \frac{g(t_1 + t_2)}{2}$

This important result states that the initial vertical velocity component ($u \sin \theta$) is related to the sum of the times $t_1$ and $t_2$. Note that $t_1 + t_2$ is equal to the total time of flight $T = \frac{2u \sin \theta}{g}$.

Substituting $u \sin \theta = \frac{g(t_1 + t_2)}{2}$ back into equation (i) to find $y$:

$y = \left(\frac{g(t_1 + t_2)}{2}\right) t_1 – \frac{1}{2} g t_1^2$
$y = \frac{g t_1^2}{2} + \frac{g t_1 t_2}{2} – \frac{1}{2} g t_1^2$
$y = \frac{g t_1 t_2}{2}$

The times $t_1$ and $t_2$ are the roots of the quadratic equation for $t$ from $y = (u \sin \theta) t – \frac{1}{2} g t^2$ when $y$ is a specific height:

$\frac{1}{2} g t^2 – (u \sin \theta) t + y = 0$
$t^2 – \frac{2u \sin \theta}{g} t + \frac{2y}{g} = 0$

Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$:

$t = \frac{\frac{2u \sin \theta}{g} \pm \sqrt{\left(\frac{2u \sin \theta}{g}\right)^2 – 4 \times 1 \times \frac{2y}{g}}}{2 \times 1}$
$t = \frac{u \sin \theta}{g} \pm \frac{1}{2} \sqrt{\frac{4u^2 \sin^2 \theta}{g^2} – \frac{8y}{g}}$
$t = \frac{u \sin \theta}{g} \left[1 \pm \sqrt{1 – \frac{8yg^2}{4u^2 \sin^2 \theta g}}\right]$
$t = \frac{u \sin \theta}{g} \left[1 \pm \sqrt{1 – \left(\frac{\sqrt{2gy}}{u \sin \theta}\right)^2}\right]$

So, $t_1$ and $t_2$ are:

$t_1 = \frac{u \sin \theta}{g} \left[1 + \sqrt{1 – \left(\frac{\sqrt{2gy}}{u \sin \theta}\right)^2}\right]$
$t_2 = \frac{u \sin \theta}{g} \left[1 – \sqrt{1 – \left(\frac{\sqrt{2gy}}{u \sin \theta}\right)^2}\right]$

Consider two balls, A and B, projected simultaneously from the origin with initial velocities $\vec{u}_1$ and $\vec{u}_2$ at angles $\theta_1$ and $\theta_2$ respectively with the horizontal.

The instantaneous positions of the two balls at time $t$ are given by:

For Ball A:

$x_1 = (u_1 \cos \theta_1) t$
$y_1 = (u_1 \sin \theta_1) t – \frac{1}{2} g t^2$

For Ball B:

$x_2 = (u_2 \cos \theta_2) t$
$y_2 = (u_2 \sin \theta_2) t – \frac{1}{2} g t^2$

The position of ball A with respect to ball B ($\vec{r}_{A/B}$) is given by the relative coordinates:

$\vec{r}_{A/B} = \vec{r}_A – \vec{r}_B = (x_1 – x_2) \hat{i} + (y_1 – y_2) \hat{j}$
$x_{A/B} = x_1 – x_2 = (u_1 \cos \theta_1 – u_2 \cos \theta_2) t$
$y_{A/B} = y_1 – y_2 = ((u_1 \sin \theta_1) t – \frac{1}{2} g t^2) – ((u_2 \sin \theta_2) t – \frac{1}{2} g t^2)$
$y_{A/B} = (u_1 \sin \theta_1 – u_2 \sin \theta_2) t$

Now, let’s find the relationship between $y_{A/B}$ and $x_{A/B}$:

$\frac{y_{A/B}}{x_{A/B}} = \frac{(u_1 \sin \theta_1 – u_2 \sin \theta_2) t}{(u_1 \cos \theta_1 – u_2 \cos \theta_2) t}$
$\frac{y_{A/B}}{x_{A/B}} = \frac{u_1 \sin \theta_1 – u_2 \sin \theta_2}{u_1 \cos \theta_1 – u_2 \cos \theta_2} = \text{constant}$

Key Conclusion:

Since the ratio $y_{A/B} / x_{A/B}$ is a constant, it implies that $y_{A/B} = (\text{constant}) \times x_{A/B}$. This is the equation of a straight line passing through the origin (or point of common projection). Thus, the motion of a projectile relative to another projectile is a straight line.

Reason for Straight Line Relative Motion:

The crucial aspect is the relative acceleration. The acceleration of particle 1 ($\vec{a}_1$) is $-g\hat{j}$ and the acceleration of particle 2 ($\vec{a}_2$) is also $-g\hat{j}$. Therefore, the relative acceleration is $\vec{a}_{12} = \vec{a}_1 – \vec{a}_2 = -g\hat{j} – (-g\hat{j}) = \vec{0}$. Since the relative acceleration is zero, the relative velocity remains constant, leading to a straight-line relative path.

When a projectile moves upward, its kinetic energy decreases, and its potential energy increases. However, the total mechanical energy (Kinetic Energy + Potential Energy) always remains constant, in accordance with the law of conservation of energy (assuming no air resistance or other non-conservative forces).

If a body is projected with initial kinetic energy $K = \frac{1}{2} m u^2$ at an angle of projection $\theta$ with the horizontal, let’s analyze its energy at the highest point of the trajectory:

At the highest point:

• The vertical component of velocity is zero ($v_y = 0$).
• The horizontal component of velocity remains constant ($v_x = u \cos \theta$).
• The velocity at the highest point is $v = u \cos \theta$.
• The height is the maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$.

1. Kinetic Energy ($K’$) at the highest point:

$K’ = \frac{1}{2} m v^2 = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta$

Since the initial kinetic energy $K = \frac{1}{2} m u^2$, we can write:

$K’ = K \cos^2 \theta$

2. Potential Energy (PE) at the highest point:

$\text{PE} = mgH = mg \left(\frac{u^2 \sin^2 \theta}{2g}\right)$
$\text{PE} = \frac{1}{2} m u^2 \sin^2 \theta$

3. Total Energy (TE) at the highest point:

$\text{TE} = K’ + \text{PE}$
$\text{TE} = \frac{1}{2} m u^2 \cos^2 \theta + \frac{1}{2} m u^2 \sin^2 \theta$
$\text{TE} = \frac{1}{2} m u^2 (\cos^2 \theta + \sin^2 \theta)$
$\text{TE} = \frac{1}{2} m u^2 \times 1$
$\text{TE} = \frac{1}{2} m u^2$

This total energy is equal to the initial kinetic energy at the point of projection, demonstrating the conservation of mechanical energy.

A body projected horizontally from a certain height ‘$y’$ vertically above the ground with initial velocity ‘$u$’ undergoes horizontal projectile motion. If air friction is negligible, there is no horizontal force to affect the horizontal motion, so the horizontal velocity remains constant. The object therefore covers equal horizontal distances in equal intervals of time.

7.1. Trajectory of Horizontal Projectile (Derivation)

Consider the point of projection as the origin $(0,0)$. The initial velocity is entirely horizontal, so $u_x = u$ and $u_y = 0$. The acceleration is $a_x = 0$ and $a_y = -g$ (taking downward as negative y).

For horizontal motion:

$x = u_x t = u t \quad \Rightarrow \quad t = \frac{x}{u} \quad \cdots (i)$

For vertical motion:

$y = u_y t + \frac{1}{2} a_y t^2 = (0) t + \frac{1}{2} (-g) t^2 = -\frac{1}{2} g t^2$

If we take the downward direction as positive for the Y-axis (as in the diagram), then $y = \frac{1}{2} g t^2 \quad \cdots (ii)$

Substituting $t$ from (i) into (ii):

$y = \frac{1}{2} g \left(\frac{x}{u}\right)^2$
$y = \frac{g x^2}{2 u^2}$

This equation is of the form $x^2 = Cy$ (or $y=Cx^2$), which represents a parabola. Hence, the trajectory of a horizontal projectile is parabolic.

After time $t$, the horizontal displacement is $x = ut$ and the vertical displacement (downwards from projection point) is $y = \frac{1}{2} g t^2$.

The position vector $\vec{r}$ (from the origin, taking downwards as positive Y) is:

$\vec{r} = x \hat{i} + y \hat{j} = u t \hat{i} + \frac{1}{2} g t^2 \hat{j}$

The magnitude of the displacement $r$ is:

$r = \sqrt{(ut)^2 + \left(\frac{1}{2} g t^2\right)^2} = ut \sqrt{1 + \left(\frac{g t}{2u}\right)^2}$

The angle $\alpha$ that the displacement vector makes with the horizontal (downwards) is:

$\tan \alpha = \frac{y}{x} = \frac{\frac{1}{2} g t^2}{ut} = \frac{g t}{2u}$
Since $t = \sqrt{\frac{2y}{g}}$, we can also write:
$\tan \alpha = \frac{g}{2u} \sqrt{\frac{2y}{g}} = \frac{1}{u} \sqrt{\frac{g^2 \times 2y}{g^2}} = \frac{1}{u} \sqrt{2gy}$
$\alpha = \tan^{-1}\left(\frac{\sqrt{2gy}}{u}\right)$

Throughout the motion, the horizontal component of velocity is constant:

$v_x = u$

The vertical component of velocity increases with time due to gravity ($v = u + at$):

$v_y = u_y + a_y t = 0 + g t = g t$ (taking downward as positive y)

The instantaneous velocity vector $\vec{v}$ is:

$\vec{v} = v_x \hat{i} + v_y \hat{j} = u \hat{i} + g t \hat{j}$

The magnitude of the instantaneous velocity $v$ is:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + (gt)^2} = u \sqrt{1 + \left(\frac{gt}{u}\right)^2}$

Alternatively, since $t = \sqrt{\frac{2y}{g}}$ for vertical fall, $gt = \sqrt{2gy}$. So:

$v = \sqrt{u^2 + 2gy}$

Direction of instantaneous velocity (angle $\phi$ below the horizontal):

$\tan \phi = \frac{v_y}{v_x} = \frac{gt}{u}$
$\phi = \tan^{-1}\left(\frac{gt}{u}\right) = \tan^{-1}\left(\frac{\sqrt{2gy}}{u}\right)$

If a body is projected horizontally from a height $h$ with initial velocity $u$, the time taken for the body to reach the ground is determined solely by its vertical motion.

Using the vertical motion equation ($y = u_y t + \frac{1}{2} a_y t^2$), with $u_y = 0$, $a_y = g$, and vertical displacement $h$:

$h = 0 \cdot T + \frac{1}{2} g T^2$
$h = \frac{1}{2} g T^2$
$T^2 = \frac{2h}{g}$
$T = \sqrt{\frac{2h}{g}}$

Key Implication:

The time of flight for a horizontally projected body depends only on the height from which it is projected and the acceleration due to gravity, NOT on its initial horizontal velocity. This means if multiple particles are released/projected horizontally from the same height, they will all hit the ground simultaneously.

The horizontal range $R$ is the horizontal distance traveled by the body during its time of flight $T$. Since the horizontal velocity is constant ($u_x = u$):

$R = u_x \times T = u T$

Substituting the time of flight $T = \sqrt{\frac{2h}{g}}$:

$R = u \sqrt{\frac{2h}{g}}$

This type of projectile motion involves a particle projected from an inclined plane, or on an inclined plane. The key difference is that the acceleration due to gravity ($g$) now has components both parallel and perpendicular to the inclined plane, affecting both the “horizontal” (along the incline) and “vertical” (perpendicular to the incline) motions.

Let the inclined plane make an angle $\alpha$ with the horizontal. Let the projectile be thrown with a speed $u$ at an angle $\theta$ with respect to the inclined plane.

Choosing a coordinate system:

It is convenient to choose the X-axis along the inclined plane (upwards) and the Y-axis perpendicular to the inclined plane (upwards).

Components of initial velocity:

• Component parallel to the plane ($u_{||}$ or $u_x$): $u_x = u \cos \theta$
• Component perpendicular to the plane ($u_{\perp}$ or $u_y$): $u_y = u \sin \theta$

Components of acceleration due to gravity ($g$):

The acceleration due to gravity $g$ acts vertically downwards. Its components along the chosen axes are:

• Component parallel to the plane ($a_{||}$ or $a_x$): This component acts downwards along the plane.

  $a_x = -g \sin \alpha$ (negative because it’s opposing the upward motion along the incline)

• Component perpendicular to the plane ($a_{\perp}$ or $a_y$): This component acts downwards perpendicular to the plane.

  $a_y = -g \cos \alpha$ (negative because it’s opposing the upward motion perpendicular to the incline)

Thus, the particle decelerates at a rate of $g \sin \alpha$ as it moves up the incline.

The time of flight is the total time for which the projectile remains in the air, from the point of projection until it lands back on the inclined plane. This is determined by the vertical motion (perpendicular to the plane).

Using the second equation of motion for vertical motion ($y = u_y t + \frac{1}{2} a_y t^2$). When the projectile lands back on the incline, its vertical displacement ($y$) with respect to the incline is zero.

$0 = (u \sin \theta) T + \frac{1}{2} (-g \cos \alpha) T^2$
$0 = T \left(u \sin \theta – \frac{1}{2} g \cos \alpha T\right)$

Two solutions: $T=0$ (at projection) or $u \sin \theta – \frac{1}{2} g \cos \alpha T = 0$.

$\frac{1}{2} g \cos \alpha T = u \sin \theta$
$T = \frac{2 u \sin \theta}{g \cos \alpha}$

Note:

This formula is analogous to the time of flight for oblique projectile motion on horizontal ground ($T = \frac{2u \sin \theta}{g}$), where $u_{\perp}$ (component of $u$ perpendicular to trajectory) is $u \sin \theta$ and $a_{\perp}$ (component of $g$ perpendicular to trajectory) is $g \cos \alpha$. Thus, $T = \frac{2u_{\perp}}{a_{\perp}}$.

The maximum height is measured perpendicular to the inclined plane. At the maximum height, the component of velocity perpendicular to the plane becomes zero ($v_y = 0$).

Using the third equation of motion for vertical motion ($v_y^2 = u_y^2 + 2 a_y H$).

$0^2 = (u \sin \theta)^2 + 2 (-g \cos \alpha) H$
$0 = u^2 \sin^2 \theta – 2g \cos \alpha H$
$2g \cos \alpha H = u^2 \sin^2 \theta$
$H = \frac{u^2 \sin^2 \theta}{2g \cos \alpha}$

Note:

This formula is analogous to the maximum height for oblique projectile motion on horizontal ground ($H = \frac{u^2 \sin^2 \theta}{2g}$), where $u_{\perp}$ is $u \sin \theta$ and $a_{\perp}$ is $g \cos \alpha$. Thus, $H = \frac{u_{\perp}^2}{2a_{\perp}}$.

The horizontal range $R$ is the distance covered along the inclined plane during the time of flight $T$.

Using the second equation of motion for motion parallel to the plane ($s = u_x t + \frac{1}{2} a_x t^2$):

$R = (u \cos \theta) T + \frac{1}{2} (-g \sin \alpha) T^2$
$R = u \cos \theta T – \frac{1}{2} g \sin \alpha T^2$

Substitute $T = \frac{2 u \sin \theta}{g \cos \alpha}$ into the equation for $R$:

$R = (u \cos \theta) \left(\frac{2 u \sin \theta}{g \cos \alpha}\right) – \frac{1}{2} g \sin \alpha \left(\frac{2 u \sin \theta}{g \cos \alpha}\right)^2$
$R = \frac{2 u^2 \sin \theta \cos \theta}{g \cos \alpha} – \frac{1}{2} g \sin \alpha \frac{4 u^2 \sin^2 \theta}{g^2 \cos^2 \alpha}$
$R = \frac{2 u^2 \sin \theta \cos \theta}{g \cos \alpha} – \frac{2 u^2 \sin \alpha \sin^2 \theta}{g \cos^2 \alpha}$
$R = \frac{2 u^2 \sin \theta}{g \cos^2 \alpha} (\cos \theta \cos \alpha – \sin \alpha \sin \theta)$
Using the identity $\cos(A+B) = \cos A \cos B – \sin A \sin B$:
$R = \frac{2 u^2 \sin \theta \cos(\theta + \alpha)}{g \cos^2 \alpha}$

The maximum range along the inclined plane is achieved when the angle of projection $\theta$ (relative to the incline) is optimal.

The range is $R = \frac{2 u^2 \sin \theta \cos(\theta + \alpha)}{g \cos^2 \alpha}$. To maximize $R$, we need to maximize $\sin \theta \cos(\theta + \alpha)$.

Using the product-to-sum identity: $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$

$2 \sin \theta \cos(\theta + \alpha) = \sin(\theta + \theta + \alpha) + \sin(\theta – (\theta + \alpha))$
$= \sin(2\theta + \alpha) + \sin(-\alpha)$
$= \sin(2\theta + \alpha) – \sin \alpha$

For $R$ to be maximum, $\sin(2\theta + \alpha)$ must be maximum, i.e., $\sin(2\theta + \alpha) = 1$.

$2\theta + \alpha = 90^\circ = \frac{\pi}{2}$ radians
$2\theta = 90^\circ – \alpha$
$\theta = 45^\circ – \frac{\alpha}{2}$

Maximum range ($R_{max}$) when projected upwards along the incline:

$R_{max} = \frac{u^2}{g(1 + \sin \alpha)}$ (This occurs when the trajectory is perpendicular to the line joining the point of projection and the farthest point on the incline.)

Maximum range ($R_{max}$) when projected downwards along the incline:

$R_{max} = \frac{u^2}{g(1 – \sin \alpha)}$