A massive vertical wall is approaching a man at a speed u . When it is at a distance of 10   m , the…

A massive vertical wall is approaching a man at a speed u. When it is at a distance of 10 m, the man throws a ball with speed 10 m/s at an angle of 37°, which after having a completely elastic collision with the wall, reaches back directly into the hands of the man. The velocity of the wall is

  1. 13 3 m/s
  2. 18 2 m/s
  3. 26 4 m/s
  4. 31 5 m/s

Solution

As the vertical motion remains unaffected

​ T = 2 u sin θ g ⇒ T = 2 × 10 × 3/5 10

T=65s    ...(1)

from coefficient of restitutation along the line of collision

e = relative velocity of separation after collision relative velocity of approach before collision



1 = v - u 8 + u

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

by the time (t1) ball collide with wall, distance travelled by man = 10 - ut1

t 1 = 10 8 + u ...(3)

by the time t 2 ball reach the hand of man t 2 = 10 - u 1 v ...(4)

Since T = t1 + t2          ...(5)
from equation (2), (3), (4), (5)                                 

u = 13 3 m/s

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