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: b or c
Explanation: on khan academy
Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring
Energy lost due to friction
So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy
v = 1.40 m/sec
Answer:
t = 0.319 s
Explanation:
With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by
v = √T/λ
Linear density is
λ = m / L
λ = 4/20
λ = 0.2 kg / m
The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg
T = W = mg
T = 80 9.8
T = 784 N
The pulse rate is
v = √(784 / 0.2)
v = 62.6 m / s
The time it takes to reach the hook can be searched with kinematics
v = x / t
t = x / v
t = 20 / 62.6
t = 0.319 s
