Question

A point mass is shot vertically up from ground level with a velocity of 4 m/s at time, t = 0. It loses 20% of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is $10{\text{m/s}}^2$ and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of

• A. 1
• B. 2
• C. 4
• D. $\infty$

Answer 1

The correct option is C 4


$\left(1\right)\to t=?$
$v=u+at$
$0=4-10t$
$t=\frac{4}{10}=0.4s$
$\left(2\right)\to t^{\prime }=?$
$u^{\prime }=0.8\times u=0.8\times 4=3.2$ m/s
$v^{\prime }=u^{\prime }+a{t}^{\prime }$
$0=3.2-10t^{\prime }$
$t^{\prime }=\frac{3.2}{10}=0.32s$
$\left(3\right)\to t^{\prime \prime }=?$
$u^{\prime \prime }=0.8u^{\prime }=0.8\times 3.2=2.56$ m/s
$v^{\prime \prime }=u^{\prime \prime }+a{t}^{\prime \prime }$
$0=2.56-10t^{\prime \prime }$
$t^{\prime \prime }=0.256s$
So, $t,t^{\prime },t^{\prime \prime }$are forming a GP series
So, total time$=2(t+t^{\prime }+t^{\prime \prime }+...0)$
$=2[0.4+0.32+0.256+...0]$
$=2\times \frac{0.4}{1-0.8}=2\times 2=4s$