Question

A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far away does the book land?

Answer 1

2.45 m

Step-by-step explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

`h=(1)/(2)gt^2`

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,

`t=\sqrt{(2h)/(g)}=\sqrt{(2(1.19))/(9.8)}=0.49 s`

Now we can find the horizontal distance covered by the book, which is given by

`d=v_x t`

where

`v_x = 5.0 m/s` is the horizontal velocity

t = 0.49 s is the time of flight

Substituting,

`d=(5.0)(0.49)=2.45 m`

So the book lands 2.45 m away.