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 (Fig. E3.19). If you toss the coin with a velocity of 6.4 m/s at an angle of $\mathbf{60}\mathbf{°}$above the horizontal, the coin will land in the dish. Ignore air resistance.

(a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

The given data can be listed below,

• The velocity of coin is,$v_0=6.4m/s$
• The coin is tossed at an angle of$\theta =60°$ .
• Distance of shelf from quarter is$x=2.1m$ .

When an item (a projectile) is launched toward the surface of the Earth and deflected solely by gravity, it moves in both the horizontal and vertical planes along a curved path. This motion is said to be projectile motion.

The time of flight is given by the equation of motion in horizontal direction

$x=x_0+x_{0y}t+\frac{1}{2}{a}_x{t}^2$

Here,$x_0$ is the initial position of quarter,$v_{0x}$ is the horizontal component of initial velocity, t is the flight time, $a_x$is the acceleration in horizontal direction whose value is zero.

Substitute all the values in the above,

$$\begin{gathered} 2.1m=0+\left(6.4m/s\right)\cos \theta t+0 \\ t=\frac{2.1m}{\left(6.4m/s\right)\cos 60°} \\ t=0.656s \end{gathered}$$

The direction of velocity of the quarter is shown in the diagram below as,

The height of the shelf is given by,

$\left(H-h\right)=v_{oy}t-\frac{1}{2}g{t}^2$

Here,$v_{0y}$ is the vertical component of initial velocity, t is the flight time, -gis the acceleration in vertical direction.

Substitute all the values in the above,

$$\begin{gathered} \left(H-h\right)=v_0\sin 60°\left(0.656s\right)-\left(0.5\right)\left(9.8m/s^2\right){\left(0.656s\right)}^2 \\ =5.54m/s\left(0.656s\right)-4.9m/s^2{\left(0.656s\right)}^2 \\ =1.52m \end{gathered}$$

Thus, the height of the shelf is 1.52 m .

The vertical component of final velocity of the quarter is given by,

$v_y=v_{oy}-gt$

Here,$v_{0y}$ is the vertical component of initial velocity, t is the flight time, -gis the acceleration in vertical direction.

Substitute all the values in the above equation.

$$\begin{gathered} v_y=v_0\sin 60°-\left(9.8m/s^2\right)\left(0.656s\right) \\ =5.54m/s-6.42m/s \\ =-0.89m/s \end{gathered}$$

Thus, the vertical component of the final velocity of shelf is -0.89 m/s .