V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
Answer:
(A) No
(B) Speed decreases
Explanation:
(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.
(B) Since the person had not velocity in the horizontal direction before landing on the boat but now has one after landing on the boat, the speed of the boat will decrease because the momentum has to be conserved (remember there is no change in it).
I believe there should be some sort of table attached. Unfortunately I cannot answer this question. Sorry!
