Answer to Question #200508 in Mechanics | Relativity for Mae

Question

Students in the lab (see Figure 10.5) measure the speed

of a steel ball to be 8.0 m/s when launched horizontally

from a 1.0-m-high tabletop. Their objective is to place a

20-cm-tall coffee can on the floor to catch the ball. Show

that they score a bull’s-eye when the can is placed 3.2 m

from the base of the table.

Answer 1

Equations for coordinates of the ball:

vertical:

$y=y_0-gt^2/2$

horizontal:

$x=vt$

We have:

$y_0=1\ m$

$y=0.2\ m$

$v=8\ m/s$

$x$ is distance between the base of the table and coffee can.

Then:

$t=\sqrt{2(y_0-y)/g}=\sqrt{2(1-0.2)/9.8}=0.4\ s$

$x=8\cdot0.4=3.2\ m$