Not Phyisics???!?

Question

A marble rolls off the edge of a table that is 0.734 m high. The marble is moving at a speed of 0.122 m/s at the moment that it leaves the edge of the table. How far from the table does the marble land?

Can someone please explain how to do this please?!?! Thank you.

Answer 1

Projectile

intial vertical component =0

s=1/2at^2+Ut
U=0
S=0.734
a=9.8
0.734=1/2(9.8)(T)^2
t=0.387 s.

How far from edge of table
Disatnce = 0.122*T
0.122*(0.387)=0.0472 m.

Answer 2

This is a kinetmatics problem in two dimensions.

The horizontal equation of motion is

x - xo = vox t

x is the final horizontal position of the marble (what you're looking for)
xo is the initial horizontal position (zero)
vox is the initial horizontal velocity (.122 m/s)
t is time (which you don't yet know)

There are two unknowns so you can't solve this equation, yet. So let's look at the vertical dimension.

The vertical equation of motion is

y - yo = voy t - 1/2 g t^2

y is the final vertical position of the marble (zero)
yo is the initial vertical position of the marble (.734 m)
voy is the initial vertical velocity (zero)
g is the local acceleration of gravity, 9.8 m/s^2

Solve the second equation for t (time), then use that time in the first equation to find the final horizontal position.

This is a typical strategy. The equations of motion in each dimension are independent except for time, so use one dimesion to find the time and then carry that answer into the other dimension to finish the problem.