In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point. If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin will land in the dish. Ignore air resistance. What is the height of the shelf above the point where the quarter leaves your hand?

what is the height of the shelf above the point where the quarter leaves your hand

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 $1.53\space meters$.


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:

\[U_y = V_y -gt\]


$V_y$ 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 $-0.93 \space m/s$.

Previous Question < > Next Question