91Ó°ÊÓ

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$\mathrm{θ}=60°$ .
• 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,

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

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 .