Respuesta :
Answer:
A. [tex]y_{o}=1.17m[/tex]
B.[tex]x=0.93m[/tex]
C. [tex]v_{x}=1.9m/s[/tex]
D. [tex]v_{y}=-4.8m/s[/tex]
E. [tex]v=5.16m/s[/tex]
F.[tex]\alpha =-68.4º[/tex]
Explanation:
From the exercise we know the initial x-component of the book's velocity and how long does it take to strike the floor
[tex]v_{ox}=1.9m/s\\t=0.49s[/tex]
A. To calculate the height of the table top above the floor we need to use the equation of position from free falling objects
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
When the book strikes the floor y=0 and its initial velocity in the y component is 0 because is the book is sliding from an horizontal table
[tex]0=y_{o}-\frac{1}{2}(9.8m/s^2)(0.49s)^2[/tex]
[tex]y_{o}= \frac{1}{2}(9.8m/s^2)(0.49s)^2=1.17m[/tex]
B. To calculate the magnitude of the horizontal component we need to analyze the x component
[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^2=0+v_{ox}t+0[/tex]
[tex]x=(1.9m/s)(0.49s)=0.93m[/tex]
C. To find the magnitude of the horizontal component of the book's velocity we need to solve the following equation:
[tex]v_{x}=v_{ox}+at=1.9m/s+0=1.9m/s[/tex]
D. To find the magnitude of the vertical component of the book's velocity we need to solve the following equation:
[tex]v_{y}=v_{oy}+gt=0+(-9.8m/s^2)(0.49s)=-4.8m/s[/tex]
E. To calculate the resultant of the two components we need to add the two vectors:
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(1.9m/s)^{2}+(-4.8m/s)^{2}}=5.16m/s[/tex]
F. To find the direction of the book's velocity we need to calculate the arctan
[tex]\alpha=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-4.8m/s}{1.9m/s} )=-68.4º[/tex]
Velocity is the rate of change of position. The direction of the book's velocity just before the book reaches the floor -68.4°.
What is Velocity?
The rate of change of position of an object with respect to time is known as velocity.
As it is given to us that the initial velocity of the book is 1.9 m/s while the time taken to strike the bottom is 0.49 sec.
A.) The height of the tabletop above the floor.
In order to calculate the height of the tabletop above the floor, we will use the second equation of motion for the free-falling objects, also, we know that when the book hits the floor, its vertical distance from the ground is 0, while the initial velocity in the vertical direction is 0 m/s. therefore, the equation can be written as,
[tex]y = y_o + v_{oy} + \dfrac{1}{2}gt^2\\\\0 = y_o + 0 - \dfrac{1}{2}(9.81)(0.49)^2\\\\y_o = 1.17\rm\ m[/tex]
B.) The horizontal distance from the edge of the table to the point where the book strikes the floor.
In order to find the horizontal distance from the edge of the table to the point where the book strikes the floor, we will analyze the x-component,
[tex]x= x_o + v_{ox}t + \dfrac{1}{2}at^2\\\\x = 0 + v_{ox}t + 0\\\\x = v_{ox}t\\\\x =1.9 \times 0.49 = 0.93\rm\ m[/tex]
C.) The magnitude of the horizontal component of the book's velocity just before the book reaches the floor.
In order to find the magnitude of the horizontal component of the book's velocity, we will calculate the horizontal component of the final velocity of the book,
[tex]v_x = v_{ox}+at\\\\v_x = 1.9 + 0\\\\v_x = 1.9 m/s[/tex]
D.) The magnitude of the vertical component of the book's velocity just before the book reaches the floor.
In order to find the magnitude of the horizontal component of the book's velocity, we will calculate the vertical component of the final velocity of the book,
v_y = v_{oy}+gt
v_y = 0+(-9.81 \times 0.49)
v_y = -4.8 m/s
E.) The magnitude of the book's velocity just before the book reaches the floor.
In order to calculate the results of the two components we need to add both the components of the velocity,
[tex]v = \sqrt{v_x^2 + v_y^2}\\\\v = \sqrt{1.9^2 + (-4.8)}\\\\v = 5.16\rm\ m/s[/tex]
F.) The direction of the book's velocity just before the book reaches the floor.
In order to find the angle of the book's velocity we need to calculate the tan,
[tex]\alpha = tan^{-1}(\dfrac{v_y}{v_x})\\\\\alpha = tan^{-1}(\dfrac{-4.8}{1.9})\\\\\alpha = -68.4^o[/tex]
Learn more about Velocity:
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