Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]
and
[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]
[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]
So on
[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]
[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
