Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Area=pi times radius^2 so
if area=4pi=pi times r^2
4 times pi=r^2 times pi
divide both sdies by pi
4=r^2
square root
2=radius since legnth cannot be negative
radius=1/2 diamters
2 radius=diameter
2(2)=diameter=4
diameter=4
My guess is he gets heads about half the time
Answer:
2.5
×
t
=
2.5
t
Step-by-step explanation:
Answer:
3 hr. 45 min.
Step-by-step explanation:
