Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.
In turn, we will resort to the application of Newton's second law.
PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

Where,
X = Desplazamiento
V = Velocity
t = Time
In this case there is no initial displacement or initial velocity, therefore

Clearing for time,



PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:
F = ma
Where,
m=mass
a = acceleration
Acceleration can also be written as,

Then





Negative symbol is because the force is opposite of the direction of moton.
PART C) Acceleration through kinematics equation is defined as




The gravity is equal to 0.8, then the acceleration is


Answer:
Friction
Explanation:
Friction keeps things from sliding into things according to newton's laws
The acceleration of an object in free fall is 
Explanation:
We can solve the problem by using Newton's second law of motion, which states that:

where
F is the net force on an object
m is the mass of the object
a is its acceleration
For the object in this problem, we have:
m = 9.9 kg is its mass
F = 97.02 N is the net force on the object
So, we can rearrange the equation to find its acceleration:

This value is actually known as acceleration of gravity (g), and it is the acceleration that any object in free fall has near the Earth's surface, regardless of its mass.
Learn more about Newton's second law here:
brainly.com/question/3820012
And about free fall:
brainly.com/question/1748290
brainly.com/question/11042118
brainly.com/question/2455974
brainly.com/question/2607086
#LearnwithBrainly
Answer:

Given:
Input Power (
)= 1000 W
Mass (m) = 100 kg
Height (h) = 10 m
Time (t) = 16 s
To Find:
Efficiency of the motor
Explanation:










Efficiency of the motor = 62.5 %