A basket ball is dropped from a height of 20 feet It bounces back each time to a height which

total distance covered will be 60 feet appx..distance covered in each bounce will be twice the height of that bounce.
20ft + 2*10ft(rebounce)+ 2*5ft(rebounce)+2*2.5(rebounce)+2*1.25(rebounce)+...==60ft
10 years agoHelpfull: Yes(5) No(0)
sorry i had taken height as 10
for 20 20*2*2-20
60
10 years agoHelpfull: Yes(1) No(0)
here r=1/2
therefore we kmow a/(1-r) for infinite g p series it is used so dor 1st and 2nd bounce we have
(20/(1-1/2))+10/(1-1/2)=60feet ans
10 years agoHelpfull: Yes(1) No(0)
here we cannot determine the the distance travelled becoz it wont come to rest it will always bounce half,that means suppose after nth bounce the distance is 0.0001 micro feet. so in next bounce i.e n+1 bounce it will bounce 0.0001/2 micro feet..
10 years agoHelpfull: Yes(0) No(5)
10+2*10*1/2+2*10*(1/2)^2...........
2*(10*1/2+10*1/2^2.....)-10=2*10/(1-1/2)-10
20*2-10=30
10 years agoHelpfull: Yes(0) No(2)
60, as 20+10+10+5+5+2.5+2.5+1.25+1.25+0.625+0.625+..=60ft
10 years agoHelpfull: Yes(0) No(0)
Distance covered = 20 + 10 + 10 + 5 + 5 + 2.5 + 2.5 + 1.25 + 1.25 + ...
It is a GP and we use sum of infinite no. formula = a(1/1-r)
a=20 r=1/2
= 20 + 20(1/1-(1/2))
= 20 + 40 [sum of infinite GP = a/(1-r)]
= 60 feet.
2 years agoHelpfull: Yes(0) No(0)

A basket ball is dropped from a height of 20 feet.It bounces back each time to a height which is one half of hthe height of the last bounce. How far approximately will the ball have travelled before it comes to rest?

A. 30 feet

B. 40 feet

C. 60 feet

D. Can not be determined