Q.A boy throws a ball in air at $60^\circ$ to the horizontal along a road with a speed of 10 m/s (36km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of (18km/h). Give explanation to support your diagram.

The key insight: In projectile motion, the minimum speed is at the highest point, because there the vertical velocity is zero and only the horizontal component $u_x$ remains. So the given $15\ \mathrm{m/s}$ is actually $u_x$. The maximum height $H$ tells us the vertical component $u_y$ via $u_y^2 = 2gH$. Then the projection speed is $u = \sqrt{u_x^2 + u_y^2}$.

1. Identify the minimum velocity.

   At the peak, $v_y = 0$, so the speed is just $|v_x| = u_x$. Hence $u_x = 15\ \mathrm{m/s}$.

2. Find the vertical component from maximum height.

   Using $v_y^2 = u_y^2 - 2gH$ with $v_y = 0$ at the top:

3. Combine components to get projection speed. …

Convert speeds using $1\ \text{kmph}=\tfrac{5}{18}\ \mathrm{m/s}$:

• Downstream speed $=(18+10.8)=28.8\ \text{kmph}=28.8\times\tfrac{5}{18}=8\ \mathrm{m/s}$.
• Upstream speed $=(18-10.8)=7.2\ \text{kmph}=7.2\times\tfrac{5}{18}=2\ \mathrm{m/s}$.

Times for each leg:

$t_{\text{down}}=\frac{120}{8}=15\ \text{s},\qquad t_{\text{up}}=\frac{90}{2}=45\ \text{s}.$ …

We start by recognising that the cannonball is launched from a height of $375\ \mathrm{m}$ above the ground. The initial speed is $100\ \mathrm{m/s}$ at $30^\circ$ above the horizontal. The acceleration due to gravity is $10\ \mathrm{m/s^2}$ downward.

The standard approach for projectile motion is to treat horizontal and vertical motions independently. The horizontal motion has constant velocity; the vertical motion has constant acceleration. Because the ball lands below the launch height, the time of flight is found by solving the vertical displacement equation for when the vertical position equals $-375\ \mathrm{m}$ (taking upward as positive).

1. Resolve the initial velocity into components

   Horizontal: $u_x = 100 \cos 30^\circ = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3}\ \mathrm{m/s}$

   Vertical: $u_y = 100 \sin 30^\circ = 100 \cdot \frac{1}{2} = 50\ \mathrm{m/s}$

2. Write the vertical displacement equation

   Taking upward as positive, the vertical displacement $y$ at time $t$ is

The ball starts at $y=0$ (cliff top) and lands at $y = -375\ \mathrm{m}$ (ground level below). So we set

Simplify:

Rearranging:

Divide through by 5:

3. Solve the quadratic for time of flight

The positive root is $t = \frac{10 + 20}{2} = 15\ \mathrm{s}$. (The negative root is discarded.)

4. Compute the horizontal distance …

The key idea is that projectile motion is a combination of constant horizontal velocity (no horizontal acceleration) and uniformly accelerated vertical motion (due to gravity). At the same vertical height, the vertical speed is the same magnitude but opposite direction compared to the launch point.

1. Identify the components at point A

   The initial velocity is $\vec{v}_A = 4\vec{i} + 5\vec{j}$ m/s.

   • Horizontal component: $v_x = 4$ m/s (constant throughout the flight).
   • Vertical component: $v_y = 5$ m/s upward.

2. Understand what happens at point B

   Point B is at the same height as point A (the figure shows a symmetric trajectory). In projectile motion, when the projectile returns to the same vertical level, the vertical velocity has the same magnitude but opposite direction (because gravity has acted symmetrically on the way up and down).

3. Apply the symmetry

   • Horizontal velocity unchanged: $v_x = 4$ m/s (still positive $\vec{i}$ direction).
   • Vertical velocity at B: $v_y = -5$ m/s (downward, so $-\vec{j}$). …

The key idea is that on a displacement–time graph, the slope (gradient) at any point represents the instantaneous velocity. For straight-line graphs, the slope is constant, so the velocity is simply the tangent of the angle the line makes with the time axis. Therefore, comparing velocities reduces to comparing the tangents of those angles.

1. Identify the slopes.

   The displacement–time graph of a particle moving with constant velocity is a straight line. The slope of that line is $\tan\theta$, where $\theta$ is the angle the line makes with the time axis (the positive $t$-axis).

   • For the first particle, $\theta_1 = 30^\circ$, so its velocity $v_1 = \tan 30^\circ$.
   • For the second particle, $\theta_2 = 45^\circ$, so its velocity $v_2 = \tan 45^\circ$.

2. Compute the tangents.

   • $\tan 30^\circ = \frac{1}{\sqrt{3}}$
   • $\tan 45^\circ = 1$

3. Form the ratio.

   The ratio of their velocities is

   $$$$

Concept and Intuition

When two objects move with constant velocities, the distance between them changes in a way that depends on their relative motion. The distance of closest approach occurs at the instant when the relative velocity is perpendicular to the relative position vector. Why? Because if you imagine standing on one car and watching the other, the line joining you shrinks fastest when the other car moves directly toward you; once it starts moving "sideways" relative to that line, the distance begins to increase again. The turning point — the minimum separation — happens exactly when the relative velocity has no component along the line joining them.

So instead of solving messy coordinate equations, we work in the frame of one car and use the condition $\vec{v}_{rel} \cdot \vec{r}_{rel} = 0$.

Step-by-step solution

1. Set up coordinates and initial positions

Let the south–north direction be the $y$-axis (north positive) and the west–east direction be the $x$-axis (west positive, since both cars move west or north). Place the southern car at the origin at $t=0$:

• Southern car (Car A): at $(0,0)$, moving north at $25$ km/hr.

  $\vec{v}_A = (0, 25)$ km/hr.

• Northern car (Car B): initially 50 km north of Car A, so at $(0,50)$. It moves west at $25$ km/hr.

  $\vec{v}_B = (25, 0)$ km/hr (west is positive $x$).

2. Find relative position and relative velocity

Relative position of Car B with respect to Car A:

Relative velocity of Car B with respect to Car A:

So Car B is moving, in Car A's frame, with a velocity that has equal components eastward (wait — careful: $+x$ is west, so $25$ km/hr west relative to Car A) and southward (since $-25$ in $y$ means south). The magnitude of $\vec{v}_{BA}$ is $25\sqrt{2}$ km/hr, directed southeast.

3. Condition for closest approach

At time $t$, the relative position vector is:

The distance $d(t) = |\vec{r}_{BA}(t)|$ is minimum when $\vec{v}_{BA} \cdot \vec{r}_{BA}(t) = 0$ (the relative velocity is perpendicular to the line joining them). …

When an object is launched into the air, undergoing projectile motion, its trajectory is influenced only by gravity (neglecting air resistance). This means the acceleration acts purely in the vertical direction. A key insight here is that the horizontal component of the velocity remains constant throughout the flight, while the vertical component changes uniformly due to gravity. The angle the velocity vector makes with the horizontal depends on the ratio of its vertical and horizontal components.

Here's how we can determine the time:

1. Resolve the initial velocity into components: The initial velocity $u = 140$ m/s is given at an angle of $\theta_0 = 60^\circ$ with the horizontal. We break this down into its horizontal ($u_x$) and vertical ($u_y$) components:

2. Determine velocity components at time $t$: As discussed, the horizontal velocity component remains constant because there is no horizontal acceleration.

The vertical velocity component changes due to gravity. Since the object is thrown upwards, gravity acts downwards, reducing the upward velocity.

Substituting the values:

3. Relate the velocity components to the final angle: At some time $t$, the velocity vector $\vec{v}$ makes an angle $\theta_f = 30^\circ$ with the horizontal. The tangent of this angle is the ratio of the vertical velocity component to the horizontal velocity component: $\tan \theta_f = \frac{v_y}{v_x}$ …