This problem aims to familiarize us with the **projectile motion** of an object where a coin is thrown in a dish with some **horizontal velocity**. This problem requires the concepts of** projectile motion, momentum, **and **complementary angles.**

Now, **projectile motion** is a type of motion in which an object is **thrown** or tossed into the atmosphere with only the **acceleration of gravity** acting on the object. The object is thus referred to as a **projectile**, and its horizontal path is called its **trajectory**.

When a **projectile** is underway and the **air resistance** is insignificant, the overall **momentum** is conserved in the horizontal orientation because horizontal forces tend to be 0. **Conservation of momentum** is laid out only when the total external force is 0. Thus, we can say that the **law of conservation of momentum** holds valid when evaluating systems of particles.

## Expert Answer

The first thing we are going to do is to **resolve** the **initial speed** into its **rectangular** components that are **vertical **and **horizontal** components:

Since the **vertical component** is along the $y$-axis, it becomes $V_y = Vsin \theta$

Whereas the **horizontal component** comes out to be $V_x = Vcos \theta$.

The **initial velocity** $V$ is given as $6.4 \space m/s$.

And the **angle of projectile** $\theta$ is given as $60$.

Plugging in all the values, gives us $V_x$ and $V_y$:

\[V_x = 6.4cos60 = 3.20\space m/s\]

\[V_y = 6.4sin60 = 5.54 \space m/s\]

Now, the **projectile motion** is dependent on one thing only and that is the **time** **taken** by the coin to reach the plate, which is the ratio of the **distance** to the **horizontal velocity** of the projectile, calculated as:

\[Time \space Taken = \dfrac{Horizontal \space Distance}{Horizontal \space Velocity}\]

Plugging in the values:

\[= \dfrac{2.1}{3.2}\]

\[Time \space Taken = 0.656\]

The $2^{nd}$ **equation of motion** gives the displacement of an object under a constant gravitational acceleration $g$:

\[S = ut + 0.5gt^2\]

Where $S$ is the **height or vertical distance**,

$u$ is the** initial velocity**,

And $g$ is the **acceleration due to gravity** that is $-9.8m/s$ (negative for a downward motion).

Inserting the **values** in the formula:

\[S = (5.54 \times 0.656)+(0.5 \times -9.8 \times 0.656^2)\]

\[S = 3.635 – 2.1102\]

\[S = 1.53\]

## Numerical Result

The **height of the coin** above the point where the coin leaves your hand is $1.53\space meters$.

## Example

What is the** vertical component** of the velocity of the quarter just before it lands in the dish?

**Vertical and Horizontal components** are calculated as:

\[V_x = 3.2 \space m/s \]

\[V_y = 5.5 \space m/s\]

**Time Taken** is calculated as:

\[Time \space Taken = 0.66 \space s\]

The **vertical** component of the final velocity of the quarter is:

\[U_y = V_y -gt\]

Where,

$V_y$ is $5.5 \space m/s$

$g$ is $9.8 \space m/s$

$t$ is $0.66 \space s$

**Inserting** in to the formula:

\[U_y=5.5 – (9.8t \times 0.66)\]

\[= -0.93\]

The** vertical component** of the velocity of a quarter just before it lands in the dish is $-0.93 \space m/s$.