5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground 
is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore
or
Solving for t, we get
We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):
Answer:
The mother has to sit 2.17 ft from the center on the other side of the seesaw.
Explanation:
We are trying to find the sum of torques given by the weights of mother and daughter to be zero.
If the torque of the daughter on one side of the pivoting point is given by:
5.5 ft x 63.5 lb x g = 349.25 g ft lb
we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:
d x 160.9 lb x g = 349.25 g ft lb
d = 2.17 ft
Answer:
A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.
A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.
A 50 kg person falling at 10 m/s would have the same kinetic energy.
Explanation:
hope this helps:)
