Grade 12th passMechanics

A rubber ball is dropped from a height of 5 m on a planet, where the acceleration due to gravity is not known. on bouncing it rises to 1.8 m. The ball loses its velocity by a factor of?

16/25

2/5

3/5

9/25

Yogiraj

10 Years agoGrade 12th pass

3 Answers

Riddhish Bhalodia

Approved Tutor Answer10 Years ago

The energy loss is proportional in kinetica nd potential energy

hence
we have
$\frac{1/2mv_1^2}{1/2mv_2^2} = \frac{mg'h_1}{mg'h_2} = 25/9$
$\frac{v_1^2}{v_2^2} = 25/9$
thus
$\frac{v_1}{v_2} = 5/3$

JAHNAVI

10 Years ago

s = 1/2 a * t ^2

make "a" whatever you want
let a = 1

s = 1.8 = 1/2 t^2
t^2 = 3.6

t = sqrt(3.6)

on the "down" path
s = 5 = 1/2 a * t^2 (a = 1 still)
10 = t^2
t=sqrt(10)

velocity = acc * time
v = a*t (a = 1 still)

"up" v = at = sqrt(3.6)
"down" v = at = sqrt(10)

the "factor" [ the ball loses its velocity on bouncing?]

= sqrt(3.6)/sqrt(10) = sqrt (.36) = 0.6

,,,, my method is scary, but the answer is probly right ...

good luck

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

A ball dropped from a height h on reaching the surface velocity is given by
V1= √2gh1
Let V2 be the velocity with which the ball bounces. It will attain a height h2 given by
V2= √2gh2
V2/V1 = √h2/h1
Given
h1 = 5 m h2 = 1.8 m
V2/V1 = √1.8/5 = √0.36 = 0.6
1 - V2/V1 = V1 - V2 / V1 = 1 - 0.6/1
0.4 = 2/5

Thanks and Regards