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
4m+9+5m-12=42
9m-3=42
Divide bothsides by 9.
9m/9 -3/9=-0.3=-4
m=-4
Answer:
4
Step-by-step explanation:
B = ym + y
B - y = ym
(B - y) / y = m
or
B/y - 1 = m
Answer:
21
Step-by-step explanation:
when p is 4
p² + 5 = 4² + 5
16 + 5 = 21
