Q9E - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

a) The height of the tabletop is 1.13 m . 

b) The horizontal distance from edge of the table to the point at book strike the floor is 0.528 m.

c) The x and y-component of the final velocity are 1.10 m/s and $-4.704 \mathrm{m}/\mathrm{s}$ respectively, the magnitude and direction of the velocity are $4.831 \mathrm{m}/\mathrm{s}$ and $76.8°$ respectively.

d) The graph of the $x-t, y-t, v_x-t$ and $v_y-t$ are shown in step 5.  

Step 1: Identification of given data.

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 $t=0.480 \mathrm{s}$.

Step 2: Concept/Significance of velocity.

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.

Step 3: Determination of the height of the tabletop above the floor.

(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 g and $0.480s$ 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 

Step 4: Determination of the horizontal distance from the edge of the table to the point where the book strikes the floor

(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 t in 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 $0.528 \mathrm{m}$.

Step 5: Determination of the horizontal and vertical components of the book’s velocity, and the magnitude and direction of its velocity.

(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 t is 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 t in 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 t is 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 $v_y$ 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.

Step 6: Sketch of x -t, y -t, v x - t , and v y - t graphs for the motion.

(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,