Question

question no. 30 please help me
to solve
and please give correct answer

Answer 1

nd below the solution to the asked query:At the highest point the velocity of the projectile is ucos​θnow the projectile has mass 4m so the momentum possessed by the projectile at the highest point = 4mucosθ (right direction)it breaks into 3 fragments.One of the fragment falls down so its horizontal momentum is zerothe second fragment of mass m returns to the same path, hence it must have the velocity of ucosθ in the opposite direction that is the left direction.so the momentum of the 2nd fragment will be -mucosθlet the velocity of the third mass 2m be v in right direction thus the momentum of the third particle will be 2mv​Conserving momentum in the horizontal direction:we havePi=Pf4mucosθ = -mucosθ +2mv​or 5mucosθ = 2mv​so v = (5/2)ucosθ in the right direction.If you have any more doubts just ask here onthe forum and our experts will try to help you out as soon as possible. I hope this answer is helpful to you

Answer 2

maximum hught=60m
time at 3:30
range i dont know