Question

if a ball falls from rest for 3s and breaks a pane of glass which makes it lose one third of its velocity, find the total distance it will have fallen in 4 s after being dropped. I need help please

Answer 1

So recognizing the problem, we know it's a "free fall" one, so our acceleration will be the gravity "g".
Let's first know what distance it did on the first 3s.
So we will use a kinematic equation:
\[vf^2 =vi^2+2(a)(d)\]
where vf is final velocity, vi is initial velocity, a is acceleration and d, the distance it did.
Let's solve it for "d", since it's what we want to know:
\[d=\frac{ vf^2-vi^2 }{ 2a }\]
Let's think a little, it says it starts from "rest" meaning that it starts from vi=0 and it falls, so that means a=g.
So, rewriting:
\[d=\frac{ vf^2 }{ 2g }\]
Good, now, we need to know that final velocity, it says that before crashing, it fell for 3s, that's what we want to know, so, the acceleration is the gravity, so by definition:
\[g=\frac{ vf-vi }{ tf-ti }\]
since it started from rest, we say it has v=0 and ti=0, so, it'll look like this:
\[g=\frac{ vf }{ tf }\]
Then:
\[vf=(tf)(g)\]
So replacing it on the equation:
\[d=\frac{ ((tf)(g))^2 }{ 2g }\]
And now we have found the value of "d" for the first 3 seconds, since "g" is a constant and "tf" is a known variable, we just replace:
\[d=\frac{ ((3)(9.8))^2 }{ 2(9.8) }\]
I'll leave the calculations to you.