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

Answer:

$s_x=0.40\ meters$

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.

$s_y=u_yt+\dfrac{1}{2}gt^2$

Since, $u_y=0$

$t=\sqrt{\dfrac{2s_y}{g}}$

$t=\sqrt{\dfrac{2\times 0.55}{9.8}}$

t = 0.33 s

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

$s_x=1.2\times 0.33$

$s_x=0.396\ m$

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.