a basketball is dropped from a height of 20 feet it bounces back each time to a height which is

using formula of impact,
when any object is dropped from a height of h1 .it bounces back each time to a height which is h2.the total distance covered by object(it comes to rest)
H= h1*(1+k)/(1-k), where k= squrt(h2/h1)

h1=20, h2=10 ==> total distance H= 3*20= 60 feet.
12 years agoHelpfull: Yes(20) No(2)
20 + 2(10+5+2.5+....)
apply formula of infinite G.P
a/(a+r),where r=1/2
20+2(10/1-1/2)= 60 feet
11 years agoHelpfull: Yes(15) No(3)
TOTAL DISTANCE THAT BALL HAVE TO TRAVEL IS
20+10+10+5+5+2.5+2.5+1.25+1.25+.625+.625+.......
20+20+10+5+2.5+1.25+..........
20+(20/(1-1/2))
20+40
i.e
60
11 years agoHelpfull: Yes(8) No(1)
here is a trick if ball drop for 3/4 or 1/2 or smthng like ths... then(1+2)*height.
so we get, according to d question,
(1+2)*20=60 (100 % sure)
11 years agoHelpfull: Yes(1) No(4)
it is in the form of gp in which we can use the formulae s=a/1-r+((1/2a)/1-r))
where a=20 s=1/2
11 years agoHelpfull: Yes(1) No(0)
lets assume its a series of nth gp...so first calculate in one direction...
downwards 20(1-12)=40
upwards a=10 half height
10(1-12)=20
so total travelled=60
10 years agoHelpfull: Yes(1) No(0)
cannot b determined becouse we dont know what is the lower level of the bounce ball
9 years agoHelpfull: Yes(1) No(5)
40(b)
20/3/2=13
20+13+14=40
8 years agoHelpfull: Yes(0) No(2)

a basketball is dropped from a height of 20 feet.it bounces back each time to a height which is one half of the height of the last bounce.how far approximatesly will the ball have to be travelled before it comes to rest?

A)30FT.
b)40
C)60
d)CANNOT B DETERMINED