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
Ivahew [28]
2 years ago
10

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Physics
1 answer:
Ede4ka [16]2 years ago
4 0

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

y=y_{o}+V_{oy}t-\frac{g}{2}t^{2} (1)

Where:

y=0 is the final height of the ball

y_{o}=h is the initial height of the ball

V_{oy}=V_{o}sin(0\°)=0 is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

t=0.5 s is the time at which the ball lands

g=9.8 m/s^{2} is the acceleration due gravity

So, with these conditions the equation is rewritten as:

h=\frac{g}{2}t^{2} (2)

h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2} (3)

Finally:

h=1.22 m

You might be interested in
A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 diopters in the distance-vision part,
Angelina_Jolie [31]

Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

\frac{1}{f}=-0.045

Object distance = u

Image distance = v

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m

Far point

|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

\frac{1}{f}=1.75

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m

Near point

|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m

Far point of the eye is 0.4 m

7 0
3 years ago
lightning strikes in the distance and six seconds later Thunder is heard how far away was the lightning strike
marshall27 [118]
The speed of sound really depends on the temperature and moisture in the air,
and the sound of the thunder doesn't necessarily travel straight from the lightning
to where you are.  So we can't say exactly. 

But if we use the nominal speed of 340 m/s, then 6 seconds means 2,040 meters,
or about 6,700 feet, or about 1.27 miles.
8 0
3 years ago
Question 2 - If Juan starts out at 10m/s, and in 15 s speeds up to 60 m/s, what is his acceleration?
34kurt

Answer:

B) 3.33 m/s²

Explanation:

Given that,

The initial velocity of Juan, u = 10 m/s

Final velocity of Juan,  v = 60 m/s

Time taken, t = 15 s

The acceleration is given by the relation

                          a = v-u/t  m/s²

Substituting in the above equation

                           a = 60 - 10 / 15

                           a = 3.33 m/s²

Hence, the acceleration of the Juan is, a = 3.33 m/s²

5 0
3 years ago
Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t
satela [25.4K]

1) At the top, the ball has more potential energy

2) Halfway through the fall, potential energy and kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) Potential energy at the top: 784 J

5) Potential energy halfway through the fall: 392 J

6) Kinetic energy halfway through the fall: 392 J

7) KInetic energy before hitting the ground: 784 J

Explanation:

1)

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

For the bowling ball in the problem, when it sits on top of the building it has no kinetic energy (because its speed is zero, v = 0), therefore it has more potential energy than kinetic energy.

2)

The total mechanical energy of the ball, which is the sum of the potential and the kinetic energy, is constant during the fall:

E=PE+KE=const.

When the ball is at the top, all its energy is potential energy, since the kinetic energy is zero:

E=PE=mgH

where H is the initial height.

When the ball is halfway through the fall, the height is H/2, so:

PE=mg\frac{H}{2}

which means that the potential energy is now half of the total mechanical energy: but since the total energy must be constant, this means that the kinetic energy is now also half of the total energy. Therefore, potential energy and kinetic energy are equal.

3)

When the ball is just before hitting the ground, the height of the ball is now zero

h = 0

This also means that the potential energy is zero

PE = 0

Therefore, all the energy of the ball is now kinetic energy:

KE=E

which means that the kinetic energy is maximum, and therefore it is larger than the potential energy: this is because the ball accelerates during the fall, and therefore its speed is maximum just before hitting the ground.

4)

The potential energy of the ball is given by

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

When the ball sits at the top, we have

m = 2 kg

g=9.8 m/s^2

h = 40 m

Therefore, the potential energy is

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball is given by

PE=mgh

where

m = 2 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity

When the ball is halfway through the fall, the height of the ball is

h = 20 m

Therefore, its potential energy is

PE=(2)(9.8)(20)=392 J

which is half of the initial potential energy.

6)

The kinetic energy of the ball is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the ball

v is its speed

When the ball is halfway through the fall, we have:

m = 2 kg (mass of the ball)

v = 19.8 m/s (speed of the ball)

Therefore, the kinetic energy is

KE=\frac{1}{2}(2)(19.8)^2=392 J

Which is equal to the potential energy.

7)

The kinetic energy of the ball just before hitting the ground is

KE=\frac{1}{2}mv^2

where in this case,

m = 2 kg is the mass

v = 28 m/s is the speed of the ball

Therefore, kinetic energy is

KE=\frac{1}{2}(2)(28)^2=784 J

And we see that the kinetic energy of the ball just before hitting the ground is equal to the potential energy of the ball when it sits at the top: therefore, all the mechanical energy has converted from potential energy into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0
2 years ago
What is needed to give a large boulder a large acceleration?
jok3333 [9.3K]
More force needs to be applied
6 0
2 years ago
Read 2 more answers
Other questions:
  • What is the biggest barrier to the use of renewable energy in the United States?
    5·2 answers
  • Why is glass a good substance for storing chemicals?
    7·1 answer
  • What is the weight on earth of a girl with a mass of 39 kg? N
    10·1 answer
  • A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a
    6·1 answer
  • Find the average speed of a horse that traveled east for 25km in 4 hours.
    15·1 answer
  • A 0.0625 tank contains 0.0925kg nitrogen at a gauge pressure of 5.17atm.Find the temperature of the gas in degree Celsius​
    9·1 answer
  • How are traits influenced by the environment?
    5·1 answer
  • Which one of the following best describes a fuse?
    14·1 answer
  • I MARK BRAINLIEST :D
    14·2 answers
  • What is the kinetic energy of a 10kg object that is moving with a speed of 60m/s.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!