Question

A ball is dropped from rest at the top of a 6.10 m

tallbuilding, falls straight downward and collides inelastically
withthe ground, and bounces back. The ball loses 10% of itskinetic
energy every time it collides with the ground. Howmany bounces can
the ball make and still reach a windowsill that is2.38 m above the
ground?

Answer 1

n = 5 approx

Step-by-step explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

`(v_1)/(v)` = e ( coefficient of restitution ) =
`(1)/(√(10) )`

and

`(v_1)/(v) = \sqrt{(h_1)/(6.1) }`

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

`e = \sqrt{(h_1)/(6.1) }`

`e = \sqrt{(h_2)/(h_1) }`

So on

`e^n = \sqrt{(h_1)/(6.1) }* \sqrt{(h_2)/(h_1) }*... \sqrt{(h_n)/(h_(n-1) )`

`((1)/(√(10) ))^n=(2.38)/(6.1)`= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx