Question

Plz help me guys fast..
A body of a mass 'm' falls from height 'h' on ground.If 'e' be coefficient of restitution of collision between the body and ground,then the distance travelled by body before it comes to rest is?
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Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Coffecient of restitution = e.

mass = m.

Initial height = h.

$\huge{\bold{\underline{Explanation:-}}}$

#For the formula being used refer the attachment and for figure also refer the attachment.

Total distance travelled before body comes to rest:-

$\large{\bold{S = h + 2h_1 + 2h_2 + ....}}$

$\large{S = h + 2he^2 + 2he^4 + 2he^6 ....}$

by using ${h_n = he^{2n}}$

$\large{S = h(1 + 2e^2 + 2e^4 + 2e^6 ....)}$

$\large{S = h(1 + 2e^2 [1+ e^2 + e^4 ....])}$

By using geometric progression

${[1+ e^2 + e^4 ....] = \frac{1}{1 - e^2}}$

Substituting in above equation.

$\large{S = h(1 + 2e^2) \times \frac{1}{1 - e^2}}$

Taking LCM.

$\large{S = h \frac{(1 - e^2 + 2e^2 )}{1 - e^2}}$

$\large{S = h \frac{(1 + e^2 )}{1 - e^2}}$

$\huge{\boxed{\boxed{S = h[ \frac{(1 + e^2 )}{1 - e^2}]}}}$