Q.A truck starts from rest and accelerates uniformly at $2.0\ \text{m s}^{-2}$. At $t = 10\ \text{s}$, a stone is dropped by a person standing on the top of the truck ($6\ \text{m}$ high from the ground). What are the

The key idea here is Projectile Motion. Once the stone is dropped from the truck, it acquires the truck's instantaneous velocity and then moves under the influence of gravity alone, independent of the truck's subsequent motion.

1. At $t = 10\ \text{s}$, the truck's velocity is $v_{\text{truck}} = u + at = 0 + (2.0\ \text{m s}^{-2})(10\ \text{s}) = 20\ \text{m s}^{-1}$. This becomes the initial horizontal velocity of the stone, $u_x = 20\ \text{m s}^{-1}$. Since it is "dropped", its initial vertical velocity is $u_y = 0\ \text{m s}^{-1}$.
2. The stone undergoes projectile motion for a duration of $\Delta t = 11\ \text{s} - 10\ \text{s} = 1\ \text{s}$.
3. During this $\Delta t$:
   • The horizontal velocity remains constant (neglecting air resistance): $v_x = u_x = 20\ \text{m s}^{-1}$.
   • The vertical velocity changes due to gravity: $v_y = u_y + g \Delta t = 0 + (9.8\ \text{m s}^{-2})(1\ \text{s}) = 9.8\ \text{m s}^{-1}$ (downwards).
   • The magnitude of the resultant velocity is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (9.8)^2} = \sqrt{400 + 96.04} = \sqrt{496.04} \approx 22.27\ \text{m s}^{-1}$. …

When a person drops a stone from a moving truck, the stone does not simply fall straight down relative to the ground. Instead, it carries with it the horizontal velocity of the truck at the instant it is dropped. This is a fundamental concept in relative motion and projectile motion.

Imagine you are inside a uniformly moving train and you drop a ball. From your perspective inside the train, the ball falls straight down. But to an observer standing outside the train, the ball follows a parabolic path, moving forward with the train's velocity while simultaneously falling under gravity.

In this problem, the truck is accelerating. So, the first step is to determine the truck's velocity at the precise moment the stone is dropped. This velocity then becomes the initial velocity of the stone for its subsequent projectile motion. Once the stone leaves the truck, the truck's acceleration no longer directly affects the stone. The stone is then solely under the influence of gravity (neglecting air resistance), meaning its horizontal velocity remains constant, and its vertical motion is uniformly accelerated due to gravity.

Let's break down the problem:

1. Determine the truck's velocity at $t = 10\ \text{s}$.

   The truck starts from rest, meaning its initial velocity $u_{\text{truck}} = 0\ \text{m s}^{-1}$.

   It accelerates uniformly at $a_{\text{truck}} = 2.0\ \text{m s}^{-2}$.

   We can use the first equation of motion: $v = u + at$.

   At $t = 10\ \text{s}$, the truck's velocity is:

   $v_{\text{truck}} = u_{\text{truck}} + a_{\text{truck}} t$

   $v_{\text{truck}} = 0 + (2.0\ \text{m s}^{-2})(10\ \text{s})$

   $v_{\text{truck}} = 20\ \text{m s}^{-1}$

   Important

   At the instant the stone is dropped ($t = 10\ \text{s}$), it inherits the truck's velocity. This means the initial velocity of the stone for its projectile motion is $20\ \text{m s}^{-1}$ horizontally. Its initial vertical velocity is $0\ \text{m s}^{-1}$ (relative to the ground, as it's "dropped", not thrown downwards).

2. Analyze the stone's motion from $t = 10\ \text{s}$ to $t = 11\ \text{s}$.

   The stone is dropped at $t = 10\ \text{s}$. We need its velocity and acceleration at $t = 11\ \text{s}$. This means we are interested in its state exactly $1\ \text{s}$ after it was dropped.

   Let's define a coordinate system:

   • Positive $x$-direction: direction of truck's motion.
   • Positive $y$-direction: vertically upwards.
   • The origin can be considered the point where the stone is dropped at $t=10\ \text{s}$.

   Initial velocity of the stone at the moment of dropping (let's call this $t' = 0$ for the stone's flight, corresponding to $t = 10\ \text{s}$ in the problem's timeline):

   $\vec{u}_{\text{stone}} = u_x \hat{i} + u_y \hat{j}$

   $u_x = 20\ \text{m s}^{-1}$

   $u_y = 0\ \text{m s}^{-1}$

   So, $\vec{u}_{\text{stone}} = 20\hat{i}\ \text{m s}^{-1}$.

   The acceleration acting on the stone after it is dropped is solely due to gravity, neglecting air resistance.

   $\vec{a}_{\text{stone}} = -g\hat{j}$

   Using $g = 9.8\ \text{m s}^{-2}$,

   $\vec{a}_{\text{stone}} = -9.8\hat{j}\ \text{m s}^{-2}$.

   The time duration for which the stone is in flight until $t = 11\ \text{s}$ is $\Delta t = 11\ \text{s} - 10\ \text{s} = 1\ \text{s}$.

   (a) Velocity of the stone at $t = 11\ \text{s}$:

   We need to find the velocity components at $\Delta t = 1\ \text{s}$.

   For projectile motion, the velocity components are given by:

   $v_x = u_x + a_x \Delta t$

   $v_y = u_y + a_y \Delta t$

   • Horizontal velocity ($v_x$):

     Since there are no horizontal forces (neglecting air resistance), the horizontal acceleration $a_x = 0$.

     $v_x = u_x + a_x \Delta t$

     $v_x = 20\ \text{m s}^{-1} + (0\ \text{m s}^{-2})(1\ \text{s})$

     $v_x = 20\ \text{m s}^{-1}$

   • Vertical velocity ($v_y$):

     The vertical acceleration $a_y = -g = -9.8\ \text{m s}^{-2}$.

     $v_y = u_y + a_y \Delta t$

     $v_y = 0\ \text{m s}^{-1} + (-9.8\ \text{m s}^{-2})(1\ \text{s})$

     $v_y = -9.8\ \text{m s}^{-1}$ …

Concept: A Dropped Object Inherits the Carrier's Velocity, Then Undergoes Projectile Motion

Step 1: Find the truck's velocity at the instant of drop ($t=10\ \text{s}$)

The truck starts from rest with $a_{\text{truck}}=2.0\ \text{m s}^{-2}$:

Step 2: This becomes the stone's initial velocity

When "dropped," the stone has zero vertical velocity relative to the truck but carries

the truck's horizontal velocity at that instant:

After release, the truck's acceleration no longer affects the stone — only gravity does.

Step 3: Velocity components at $t=11\ \text{s}$ (i.e. $\Delta t = 1\ \text{s}$ after release)

$v_y = u_y - g\Delta t = 0-9.8(1) = -9.8\ \text{m s}^{-1}$ …