Question

A ball is dropped from a height h above the ground . It loses 50% of its velocity on every impact with the ground before bouncing up . What is the total distance travelled by the ball when it finally comes to rest on the gound?

Answer 1

use the property of geometric series

Answer 2

Though, you have to make the assumption that eventually reducing the velocity by half each time will yield a velocity close enough to zero as to be negligible. ;-)

Answer 3

yes that's what we do.. geometric progression
h+h/2+h/2+(h/2)/2+(h/2)/2+(h/4)/2+(h/4)/2...... till infinity..
now maths is only thing left.. try to visualize it.. where ^this thing came from,

Answer 4

I think this wuestion is beyond my level

Answer 5

i have nt studied geometric progression!

Answer 6

Try choosing a value for h (like 100m) and examining a couple bounces using basic kinematics to see the pattern.

Answer 7

any other way

Answer 8

Use the first fall to find the final velocity on impact and take half of that to use as the upward velocity for the first bounce - find the new h. Do it again. Look for the pattern.
Eventually, the new h will get so small that you'll reasonably call it zero, so you stop.

Answer 9

\[ d = h \frac{( 1-(1/2)^n)}{1-1/2} = \frac{h}{1-1/2} \text { this is the easiest possible solution}\]
when n tends to infinity, (1/2)^n tends to zero ..

Answer 10

Only thing bugging me about this is that it said the velocity is reduced by half not the height.

Answer 11

i.e. half the velocity means one-forth the kinetic energy which means one-forth the potential energy which means one-forth of the height each time..

Answer 12

ya .. ya.. sorry.. v= sqrt(2gh)
v'=v/2=sqrt(2gh')=(sqrt(2gh))/2=>h'=h/4

Answer 13

The formula for each height as CliffSedge noticed is
h(n) = h0(1/2)^(2n) {1}
I checked that using kinematics.
The ball was released from initial height of h0
so the distance is
d = h0 + something
every time the ball passes through 2 h(n) of path before it rebounds from the floor.
So "something" is
\[\sum_{n=1}^{infinity} (2*h(n))\]
using {1} we get
\[d = h_{0} + \sum_{n=1}^{infinity} 2*h_{0} (1/2)^{2n} \]
or
\[d = h_{0} + 2h_{0}\sum_{n=1}^{infinity} (1/2)^{2n} \]
using wolfram we calculate the thing
http://www.wolframalpha.com/input/?i=sum+of&a=*C.sum+of-_*Calculator.dflt-&f2=+%281%2F2%29%5E%282*x%29&x=4&y=10&f=Sum.sumfunction_+%281%2F2%29%5E%282*x%29&f3=1&f=Sum.sumlowerlimit_1&f4=infinity&f=Sum.sumupperlimit_infinity&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*-

\[d = h_{0}(1+2/3)=h_{0}*5/3\]
Is the answer correct?
@openstudy1