Answer:
2.45 m
Explanation:
First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:
[tex]h=\frac{1}{2}gt^2[/tex]
where
h = 1.19 m is the vertical distance covered by the book
g = 9.8 m/s^2 is the acceleration of gravity
t is the time of flight
Solving for t,
[tex]t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s[/tex]
Now we can find the horizontal distance covered by the book, which is given by
[tex]d=v_x t[/tex]
where
[tex]v_x = 5.0 m/s[/tex] is the horizontal velocity
t = 0.49 s is the time of flight
Substituting,
[tex]d=(5.0)(0.49)=2.45 m[/tex]
So the book lands 2.45 m away.
