1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Arada [10]
3 years ago
8

A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first

rebound. (Ignore the small amount of time the ball is in contact with the floor.) (a) Determine the ball’s speed just before it hits the floor on the way down. (b) Determine the ball’s speed as it leaves the floor on its way up to its first rebound height. (c) How long is the ball in the air from the time it is dropped until the time it reaches its maximum height on the first rebound?

Physics
1 answer:
Dominik [7]3 years ago
6 0

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

You might be interested in
Which statement is true about the reaction shown by this chemical equation?
KatRina [158]

Answer:

it's C endothermic

Explanation:

5 0
2 years ago
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr
vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0
2 years ago
You perform the Hooke's Law experiment and create a plot of Displacement vs. Force. You add a linear fit and find the following
balandron [24]

Answer: 0.192 N/m

Explanation:

Well, generally when a Hooke's Law experiment is performed the plot is in fact Force vs Displacement, being the Force (in units of Newtons) in the Y-axis and the Displacement (in units of meters) in the X-axis.

In addition, if we add a linear fit the resultant equation will be the Line equation of the form:

Y=mX+b

Where m=\frac{Y_{2}-Y_{1}}{X_{2}-X_{1}} is the slope and b is the point where the line intersects the Y-axis.

So, if the equation is:

Y=0.192X+0.011

The slope of this line is 0.192 N/m which is also the spring constant k.

7 0
3 years ago
To test the performance of its tires, a car
Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

\mu mg=m\frac{v^2}{r}

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

\mu is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

g=9.8 m/s^2

And re-arranging the equation for \mu, we can find the coefficient of static friction:

\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0
3 years ago
You could use an elevator or the stairs to lift a box to tenth floor.which has greater power?
erastova [34]

Stairs don't have any power at all.  All the energy used to climb them
has to come from your muscles.

An elevator gets its power from the electric motors that lift it.  All YOU
have to do is stand there and look around.

All of this is a big part of the reason why elevators have become so
popular, and why no buildings with more than a few floors were built
before elevators were invented.


7 0
3 years ago
Other questions:
  • True or false Plant cells and animal cells have all the same major parts
    13·2 answers
  • Technician A says that magnetism can cause electric current to flow in a conductor. Technician B says that magnetic lines of flu
    11·1 answer
  • How can you calculate the number of electrons in an atom
    6·2 answers
  • If an astronaut travels to different planets, which of the following planets will the astronaut’s weight be the same as on Earth
    10·2 answers
  • A child pulls a 3.00-kg sled across level ground at constant velocity with a light rope that makes an angle 30.0° above horizont
    14·1 answer
  • Potential energy related to an object's height is called
    15·1 answer
  • While traveling on a dirt road, the bottom of a car hits a sharp rock anda small hole develops at the bottom of its gas tank. If
    11·1 answer
  • the weight of an object on the earths surface is 300N. when it is lifted to 3 times the height, its weight will become
    13·1 answer
  • 1 point
    8·1 answer
  • How many times does the light beam shown in FIGURE 26-59 reflect from (a) the top and (b) the bottom mirror?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!