A massive vertical wall is approaching a man at a speed u . When it...

As the vertical motion remains unaffected

$\text{T}=\dfrac{\text{2 u sin} \text{\theta}}{\text{g}} ⇒ \text{T}=\dfrac{2 \times \text{10} \times \text{3/5}}{\text{10}}$

$\text{T}=\dfrac{6}{5}\text{s}\text{...(1)}$

from coefficient of restitutation along the line of collision

$\text{e}=\dfrac{\text{relative velocity of separation after collision}}{\text{relative velocity of approach before collision}}$



$1=\dfrac{\text{v} - \text{u}}{8 + \text{u}}$

v - 2u=8 & & & & & & & & & & & & & & & & & & & & & & & &...(2)

by the time (t_{1}) ball collide with wall, distance travelled by man=10 - ut_{1}

${t}_{1}=\dfrac{\text{10}}{8 + \text{u}} \text{...(3)}$

$\text{by the time} \left({t}_{2}\right) \text{ball reach the hand of man} {t}_{2}=\dfrac{\text{10} - {u}_{1}}{\text{v}} \text{...(4)}$

Since T=t_{1} + t_{2} & & & &...(5)
from equation (2), (3), (4), (5)& & & & & & & & & & & & & & & & &

$\text{u}=\dfrac{\text{13}}{3} \text{m/s}$