Question

A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.20 meters/second. At what horizontal distance from the base of the table will the pen land? A.

0.18 meters
B.
0.20 meters
C.
0.40 meters
D.
0.62 meters

Answer 1

Whats the weight of the pen

Answer 2

`s_x=0.40\ meters`

Step-by-step explanation:

Given that,

Height of the table,
`s_y = 0.55 m`

Initial horizontal velocity, u = 1.2 m/s

We need to find the horizontal distance from the base of the table will the pen land. Let t is time taken by the pen to land. Let
`s_x` is the horizontal distance. It is given by :

`s_x=ut`...........(1)

Using second equation of motion to find time t.

`s_y=u_yt+(1)/(2)gt^2`

Since,
`u_y=0`

`t=\sqrt{(2s_y)/(g)}`

`t=\sqrt{(2* 0.55)/(9.8)}`

t = 0.33 s

Use the value of t in equation (1) as :

`s_x=1.2* 0.33`

`s_x=0.396\ m`

or

`s_x=0.40\ meters`

So, the horizontal distance from the base of the table 0.4 meters. Hence, this is the required solution.