A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.480 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book’s velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw x-t, y-t,${\mathit{v}}_{\mathbf{x}}\mathbf{-}\mathit{t}$, and${\mathit{v}}_{\mathbf{y}}\mathbf{-}\mathit{t}$graphs for the motion.

The given data can be listed below,

• The speed at which the book slides off is$v=1.10\mathrm{m}/\mathrm{s}$ .
• The time at which the book hits the floor is $t=0.480\mathrm{s}$.

The space at which an object's location changes in relation to a frame of reference is known as its velocity, which is a time-dependent quantity.

(a)

Assume that the initial position is at zero so the vertical displacement of the book is given by,

$y\left(t\right)=y_i+y_{i}t-\frac{1}{2}g{t}^2$

Here,$y_i$is the initial position of the book,$v_i$ is the initial velocity of the book, g is the acceleration due to gravity and t is the time at which the book falls.

Substitute 0 for $y_i$, 0 for $v_i$, 0.480s for t, $9.8\mathrm{m}/{\mathrm{s}}^2$for gand $0.480s$ for t in the above equation.

$$\begin{gathered} y=0+\left(0\right)\left(0.480\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(0.480\mathrm{s}\right)}^2 \\ =0-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right){\left(0.480\mathrm{s}\right)}^2 \\ =1.13\mathrm{m} \end{gathered}$$

Thus, the height of the tabletop is 1.13 m

(b)

The horizontal distance of the point where the book dropped and the edge of table is given by,

$x=v_{x}t$

Here,$v_x$is the velocity component of book on x-axis.

Substitute$1.10\mathrm{m}/\mathrm{s}$for${\mathrm{v}}_{\mathrm{x}}$and$0.480s$for tin the above equation.

$$\begin{gathered} x=1.10\mathrm{m}/\mathrm{s}\left(0.480\mathrm{s}\right) \\ =0.528\mathrm{m} \end{gathered}$$

Thus, the horizontal distance from edge of the table to the point at book strike the floor is $0.528\mathrm{m}$.

(c)

The y-component of final velocity of the book is given by,

$v_y=v_{0y}+a_{y}t$

Here, $v_{0y}$ is the y-component of the initial velocity of the book, $a_y$ is the y-component of acceleration and tis the time taken by the book to hit the floor.

Substitute 0m/s for $v_{0y}$, $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$and $0.480\mathrm{s}$for tin the above equation.

$$\begin{gathered} v_y=0+\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.480\mathrm{s}\right) \\ =-4.704\mathrm{m}/\mathrm{s} \end{gathered}$$

The x-component of the final velocity is given by,

$v_x=v_{0x}+a_{x}t$

Here, $v_{0x}$ is the x-component of the initial velocity of the book, $a_x$ is the x-component of acceleration and tis the time taken by the book to hit the floor.

Substitute $1.10\mathrm{m}/\mathrm{s}$for ${\mathrm{v}}_{0\mathrm{x}}$, $0\mathrm{m}/\mathrm{s}$for $a_x$and 0.480s for $t$in the above equation.

$$\begin{gathered} v_x=\left(1.10\mathrm{m}/\mathrm{s}\right)+\left(0\mathrm{m}/\mathrm{s}\right)\left(0.480\mathrm{s}\right) \\ =1.10\mathrm{m}/\mathrm{s}+0 \\ =1.10\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of velocity is given by,

$v=\sqrt{v_x^2+v_y^2}$

Substitute $1.10\mathrm{m}/\mathrm{s}$for ${\mathrm{v}}_{\mathrm{x}}$and $-4.70\mathrm{m}/\mathrm{s}$for localid="1665039820902" $v_y$in the above equation.

$$\begin{gathered} v=\sqrt{{\left(1.10\mathrm{m}/\mathrm{s}\right)}^2+{\left(-4.70\mathrm{m}/\mathrm{s}\right)}^2} \\ =\sqrt{1.21+22.127}\mathrm{m}/\mathrm{s} \\ =4.831\mathrm{m}/\mathrm{s} \end{gathered}$$

The direction of its velocity is given by,

$$\begin{gathered} \tan \alpha =\frac{v_y}{v_x} \\ \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \end{gathered}$$

Substitute $1.10\mathrm{m}/\mathrm{s}$for${\mathrm{v}}_{\mathrm{x}}$and$-4.70\mathrm{m}/\mathrm{s}$for$v_y$in the above equation.

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \\ ={\tan }^{-1}\left(\frac{-4.704}{1.10}\right) \\ ={\tan }^{-1}\left(-4.27636\right) \\ =76.8° \end{gathered}$$

Thus, the x and y-component of the final velocity are 1.10 m/s and -4.704 m/s respectively, the magnitude and direction of the velocity are 4.831 m/s and $76.8°$respectively.

(d)

The graph of the x-component of position vs time is given by,

The graph of the y-component of position vs time is given by,

The graph of the x-component of velocity vs time is given by,

The graph of the y-component of the velocity vs time is shown below,