Question

itial height at Two balls are thrown vertically from the same • Ball I is launched upward with an initial velocity voj = + 10m/s. Ball 2 is launched downward with an initial velocity vo2 = - 10m/s. same The distance between the two balls after I second from the beginning of motion is:

Answer 1

Given

vo1 = +10 m/s

vo2 = -10 m/s

Procedure

Using the free fall equations, we have:

`\begin{gathered} x1=v_(o1)t-(1)/(2)gt^2 \\ x1=10*1-(1)/(2)9.8*1 \\ x1=5.1m \end{gathered}`
`\begin{gathered} x2=v_(o2)t-(1)/(2)gt^2 \\ x2=-10*1-(1)/(2)9.8*1 \\ x2=-14.9m \end{gathered}`
`\begin{gathered} x1-x2=5.1-\lparen-14.9) \\ x1-x2=20 \end{gathered}`

The distance between the balls would be 20m