Answer:
0.91437 m
0.22859 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s² = a
When the time intervals are equal, if four drops are falling then we have 3 time intervals.
So, the time interval is
For second drop time is given by
Distance from second drop
Distance from second drop is 0.91437 m
Distance from third drop
Distance from third drop is 0.22859 m
Answer:
After throwing the object the, the velocity of the man is 13.98 m/s
Explanation:
Given:
Let,
mass of man, m1 = 74 kg
mass of box, m2 = 13 kg
Initial velocity u = 11 m/s (initially both are together hence initial velocity will be same for both)
Final velocity of man = v1
Final velocity of box = v2 = -6 m/s (the velocity is recoil velocity therefore it is negative)
To Find:
Final velocity of man,after throwing the object = v1 = ?
Solution:
Recoil velocity:
It is the backward velocity experienced.
Here recoil velocity is the backward velocity experience while throwing the box behind.Hence the velocity of the box is negative 6 m/s.
The recoil velocity is the result of conservation of linear momentum of the system. Therefore we will follow the law of conservation of momentum.
Law of conservation of momentum :
Total momentum of an isolated system before collision is always equal to total momentum after collision
substituting the values which are given above we get
Therefore, After throwing the object the, the velocity of the man is 13.98 m/s