Let from the height at which ball was thrown = h units
It is given that , ball rebounds the same percentage on each bounce.
Let it rebounds by k % after each bounce.
Height that ball attains after thrown from height h(on 1 st bounce)= [tex]h + \frac{h k}{100}=h \times (1+\frac{k}{100})[/tex]
Height that ball attains after thrown from height h (on 2 n d bounce)= [tex]h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})\times \frac{k}{100}=h \times (1+\frac{k}{100})^2[/tex]
Similarly, the pattern will form geometric sequence.
S= [tex]h +h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})^2+h \times (1+\frac{k}{100})^3+.........[/tex]
So, Common Ratio = [tex]\frac{\text{2nd term}}{\text{1 st term}}=1 +\frac{k}{100}[/tex]
Common Ratio= 1 + the percentage by which ball rebounds after each bounce
the percentage by which ball rebounds after each bounce= negative integer= k is negative integer.
Answer:
Let from the height at which ball was thrown = h units
It is given that , ball rebounds the same percentage on each bounce.
Let it rebounds by k % after each bounce.
Height that ball attains after thrown from height h(on 1 st bounce)=
Height that ball attains after thrown from height h (on 2 n d bounce)=
Similarly, the pattern will form geometric sequence.
S=
So, Common Ratio =
Common Ratio= 1 + the percentage by which ball rebounds after each bounce
the percentage by which ball rebounds after each bounce= negative integer= k is negative integer.
Step-by-step explanation:
AP3X
