A person throws a ball horizontally from the top of a building 24.0 m above the ground. this (2023)

physical middle School


answer 1

Answer: 45.24 m/s



Building height h=24 m

The range of the ball R=100 meters

Consider the vertical motion of the ball.

to use

So the initial vertical velocity is zero

Now consider the horizontal movement

, since there is no horizontal acceleration

Related questions

A weight carrier, typically weighing 400 N, stands on a very tall ladder so that it is one Earth radius above the Earth's surface (that is, twice its normal distance from the center of the earth). the earth). How much does it weigh there? answer choice group



100 newtons


First, let's find the mass of a human on the Earth's surface.

Now let's see how F changes if the distance between the two masses is doubled

Therefore, when the separation distance is doubled, the force of gravity between the two masses decreases by a quarter.

So your new weight = 400 N/4 = 100 N

Answer: It must be 100 N


A baseball with a mass of 0.685 kg is moving with a speed of 38.0 m/s when it makes contact with a baseball bat, during which time the speed of the ball changes to 57.0 m/s in the opposite direction. How much momentum does the bat transfer to the ball? When in contact with the bat, the ball experiences a maximum compression of approximately 1.0 cm.b. About how long does it take for the ball to come to rest on the bat? c) What is the average force exerted by the bat on the ball when it comes to rest?







a) Impulse sent to the ball

According to the impulse moment theorem, we have the following:



is impulsive

is the change in momentum

is the final momentum of the ball (right) with mass and final velocity

is the initial impulse (left) of the ball with initial velocity






b) time

This time can be calculated using the following equation, taking into account that the maximum compression experienced by the ball is approx.




is the acceleration

is the compressed length of the ball

is the time

From (7) it is found that:




Substitute (10) for (6):




c) The force exerted by the bat on the ball

According to Newton's second law of motion, force is proportional to the change in momentum with time:





a) 65.125 nanoseconds

b) 5,263 * 10^(-4) seconds

(c) 123737.5 Newtons


a) Impulse F.dt delivered to the ball

In terms of angular momentum, we have the following:

Using the given data, we substitute the above equation:

Answer: 65.125 nanoseconds


This time can be calculated using the following equation, taking into account that the ball is subjected to a maximum compression of approximately 0.01 meters:

Kinematic equations using constant acceleration:


vf = 0 (ball rests on bat)

vi = 38 m/s

s = compression = 0.01 m

Using the previous equation, we calculate the acceleration:

Calculate time using acceleration:

Answer: 5.263*10^(-4) seconds


According to Newton's second law of motion:

Use the answers from parts a and b

Answer: 123737.5 No

Assuming that the ship is initially at rest, immediately calculate the speed of the ship. The mass of the boy is 25.0 kg and the mass of the boat is 30.0 kg. (Take the direction of movement of the package to be positive.)



V = -0,8 m/s



De Barnett's time (m) = 25 kg

Ship mass (M) = 30 kg

ship speed = ?

Suppose a child throws a package of mass (m₁) 6 kg with a horizontal speed of 10 m/s

use conservation of momentum

(M + m + m1) V = (M+ m)V + m1 v

Plucked V = 0 m/s

(M + m + m1) x 0 = (M+ m)V + m1 v

0 = (25+50)V + 6x10

75 volts = -60

V = -0,8 m/s

Negative direction means that the speed is opposite to the direction of movement of the package.

A particle moving in a straight line slows down ???? = (−2???? 3 ) m/???? 2, where v is in m/s. If its velocity v = 10 m/s and its position s = 15 m when t = 0, determine its velocity and position at t = 25 s.


The speed of an object depends on its position.

Part A: A moving particle has a speed of 0.099 m/s at t = 25 s.

Part B: The position of the particle at t = 25 s is 15.475 m.

What is speed?

Velocity is defined as the rate at which the position of an object changes.

Since the deceleration of a particle is .. When t = 0, the velocity of the particle is v = 10 m/s and the position s = 15 m.

The acceleration of the particles is shown below.

Integrating the above equation, we get,

Muchacho t=0 s, v = 10 m/s

Part A: Speed

The velocity of the particle at t = 25 s is shown below.

Therefore, the velocity of the moving particle at t = 25 s is 0.099 m/s.

Part B: Position

The velocity of the particle is,

Integrating the above equation, we get,

Muchacho t = 0 s, s = 10 m,

Now the position of t = 25 s,

Therefore, the position of the particle at t = 25 s is 15.475 m.

To know more about speed and location, click on the link below.



(ONE). The velocity is 0.099 m/s.

(2) The position is 19.75 m.


Taking into account that,

the slowdown is

We need to calculate the speed at t = 25 s

Acceleration is the first derivative of the velocity of the particles.

About integration


Ve t = 0, v = 10 m/s

Substitute the value of C into equation (I)

The velocity is 0.099 m/s.

(2) We need to calculate the position at t = 25 seconds

Velocity is the first derivative of the particle's position.

About integration

At t = 0, s = 15 m

insert the value into the equation

The location is 19.75 m.

Therefore (a). The velocity is 0.099 m/s.

(2) The position is 19.75 m.

A block of mass 2.20 kg rests on a horizontal spring of constant k = 825 N/m and pushes the spring through 0.0650 m Questions (a) What is the elastic potential energy (J) of the block system? -spring?


Answer: 1.742 J



block masses

spring compression

The elastic potential energy of a spring-mass system is given by


A woman is standing in the ocean and notices a crest passing, and five crests passing in 50.0 seconds. The distance between two consecutive crests is 32 m. If possible, determine (a) the period of the wave.
(b) Frequency.
(c) The length of the wave.
(d) Speed.
(e) Breadth.
If any of these amounts cannot be determined, please state so.




Number of peaks (N)= 5

time (t) = 50 seconds

Distance to successive peaks = 32 m

a period of time

b) Frequency

c) The length of the wave

The distance to successive peaks is the wavelength

Wavelength = 32m

d) speed

v = f λ

v = 0,1 x 32

v = 3,2 m/s

e) Amplitude

We were unable to determine the amplitude because no data was provided.

A student saw her science teacher walking down the sidewalk next to her dorm room. She picks up a water balloon and waits. She lowered the balloon when the professor was 2 seconds directly below her window, 11 meters above the pavement. You finish the story. (assuming g # 10 m/s*)


Answer: miss the target



The distance between the window and the teacher is h=11 m

he released the balloon two seconds ago

The time it takes for the balloon to travel 11 meters

to use

where h = height

u = initial velocity

t = time

a = acceleration

So you didn't see the professor for 0.51 seconds.

A rod of mass M = 116 g and length L = 49 cm can rotate about a hinge at its left end and is initially at rest. A ball of putty of mass m = 14 g moving with velocity V = 6 m/s strikes the bar at an angle A = 37º, D = L/4 from the end and sticks to the bar after impact. (A) What is the total moment of inertia 1 of the rod-ball system about the hinge after the collision?
(B) Enter the expression for the angular velocity w of the system after the collision using m, V, D, 0,
(C) Calculate the rotational kinetic energy of the system after the collision, in joules.



a) Before:

= 0,00928 kg*m^2


= 0,00949 kg*m^2

b) W = 65,25 rad/s

c) k = 20.2021 joules


a) The moment of inertia before the collision is given by:

yo =

where M is the mass and L is the length.

yo =

= 0,00928 kg*m^2

The moment of inertia after the collision is:

yo =

The first part is the moment of inertia before the collision, but now we add the mass of the putty ball (and the distance R the putty ball hit.

yo =

= 0,00949 kg*m^2


To answer this question, we will use the law of conservation of angular momentum:

Also, the angular momentum L can be calculated by:


where M is the mass, V is the velocity, and D is the lever arm, or:

length = inside width

where I is the moment of inertia and W is the angular velocity.

According to the equation, we now have:

where is the mass of the kit ball, is the speed of the kit ball and is the moment of inertia after the collision

Replacement data:

If we solve for W, we get:

B = 65,25 rad/s


Kinetic energy is calculated using the following formula:

k =

Replacement data:

k =

k = 20.2021 joules

A motor pushes a 30.0-kg box across a wooden floor at a constant velocity of 0.75 m/s. If the coefficient of static friction from wood to wood is 0.20, what is the normal force exerted by the floor of the box?



294,3 newtons


In this case, we are told that the box does not accelerate in a horizontal plane. But neither does it accelerate in the vertical plane. This means that the sum of all vertical forces is zero.

fnet = with

Weight + Normal Force = Mass * Acceleration

-(30 kg * 9,81 m/s²) + kraft normal = 30,0 kg * 0 m/s²

Normalkraft = 294,3 N

Sound exits the diffractive horn speaker through a rectangular opening like a small door. This speaker is mounted on a pole outside. In winter, when the temperature is 273 K, the value of the diffraction angle θ is 19.5°. What is the angle of diffraction of the same sound on a summer day at 313 K?




It is well known that the speed of sound is

Now t = 273 k = 0 degrees

so, we have

Now when the temperature reaches 313 K we have

Now we have

Now from the two equations we have

so, we have

The projectile returned to its original height 4.08 s after launch, during which time it traveled 76.2 m horizontally. If air resistance can be ignored, what is the initial velocity of the projectile?


The initial velocity is the velocity of the object or the initial velocity at the beginning of the measurement.

The initial velocity of the projectile is 27.35 m/s.

What is the initial velocity?

The initial velocity is the velocity of the object or the initial velocity at the beginning of the measurement.

Information given -

The time it takes for the projectile to return to its original height is 4.08 seconds.

It covered a distance of 76.2 m.

Air resistance is negligible.

The projectile motion time can be expressed as,

Let the above formula be Formula 1.

Here is the launch velocity and the launch angle.

Rewriting the above equation as,

Let the above formula be Formula 2.

Now the range of motion of the projectile can be expressed as,

Dividing this equation by equation 1 gives,

Squaring and adding the previous equation and equation 2,

Therefore, the initial velocity of the projectile is 27.35 m/s.

Learn more about initial velocity here;


Answer: 27.35 m/s



Projectile approach time T=4.08 s

Projectile range = 76.2 m

The flight time of the projectile is given by

where u = initial velocity

launch angle

g = acceleration due to gravity

that go from

part 1 and 2

Squaring and adding 3 and 4 we get

A firework breaks into two parts, one with a 200-g mass moving along the x-axis with a velocity of 82.0 m/s, and a second 300-g mass moving along the x-axis. y-axis with a velocity of 45.0 m/s. What is the total moment of the two pieces? a) 361 kg x m/s at 56.3 degrees from the x axis.
b) 93.5 kg x m/s at 28.8 degrees from the x axis.
c) 21.2 kg x m/s at an angle of 39.5 degrees with respect to the x axis.
d) 361 kg d m/s at 0.983 degrees of the x axis.



A) 21.2 kg.m/s at an angle of 39.5 degrees with respect to the x axis


Mass of the small piece = 200g = 200/1000 = 0.2 kg

Klump Time = 300g = 300/1000 = 0.3 kg

Small part speed = 82 m/s

Speed ​​of the largest part = 45 m/s

Final moment of the small block = 0.2 × 82 = 16.4 kg.m/s

Final moment of the package = 0.3 × 45 = 13.5 kg.m/s

Since they interact at 90oC (x and y axes) and momentum is also a vector; So we can use the Pythagorean theorem

Composite moment² = 16.4² + 13.5² = 451.21

Resultant moment at 39.5 degrees from the x-axis = √451.21 = 21.2 kg.m/s (tan^-1 (13.5/16.4)

A 19 cm long bicycle arm with a pedal at one end is connected to a 23 cm diameter sprocket around which the chain moves. A cyclist on this bike increases his pedaling speed from 65 rpm to 90 rpm in 10 seconds.



The tangential acceleration of the pedal is 0.0301 m/s².


Taking into account that,

Length = 19 cm

Diameter = 23 cm

time = 10 seconds

Initial angular velocity = 65 rpm

Sluthastiged = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Use the tangential acceleration formula

Therefore, the tangential acceleration of the pedal is 0.0301 m/s².

A merry-go-round has a radius of 3.0 m and a moment of inertia of 600 kg ⋅ m2. When a 20 kg child climbs from the center to the edge, he initially rotates at 0.80 rad/s. when
The angular velocity at which the boy reaches the edge of the merry-go-round is: ________
A) 0,61 rad/s
B) 0,73 rad/s
C) 0,80 rad/s
D) 0,89 rad/s
E) 1.1 rad/s



end Angular Velocity,


Taking into account that,

Carousel radius, r = 3 m

The inertia of the carousel,

angular velocity,

Time, m = 20 kg

Let me be the new moment of inertia of the carousel. Here the angular momentum of the system is conserved. so,

Therefore, the angular velocity of the merry-go-round is 0.61 rad/s. Therefore, the correct option is (A). So this is the desired solution.

A merry-go-round with a radius of 1.80 m has a mass of 120 kg and rotates with an angular speed of 0.400 rev/s. What is its angular velocity after a 37.5 kg child grasps its outer edge and climbs? The child is initially at rest.


Answer: 0.304 rev/s




many carousels

many children

The initial moment of inertia of the system.

Assume that all the masses are at the periphery.

last moment of inertia

conservation of angular momentum

A wave in a swimming pool propagates at a speed of 0.720 m/s. You splash water at one end of the pool and watch the wave reach the other end, reflect, and return 31.0 seconds later. How far is the other end of the pool?



11,16 metros


Total distance traveled by the wave = speed * time

= 0.720 m/s * 31.0 seconds

= 22,32 metros

Since this is an echo back to the starting point, the length of the pool should be half the distance traveled.

Length of the pool = 22.32 m÷2 = 11.16 m

The stamping machine began to produce components that were out of tolerance. The manager takes the machine out of service to ensure that no more defective parts are produced and notifies maintenance personnel to repair the machine. This is an example of ________. A. Immediate Corrective Actions B. Benchmarking C. Corporate Governance D. Disciplinary Actions E. Basic Corrective Actions



Option A


Instant corrective action occurs in an instant as it triggers a response based on the specific situation.

The time period for dissolution follows the same pattern and is not sustainable.

So, a situation where a punch press breaks down and the machine is immediately taken out of service to reduce the production of defective parts is an example of immediate corrective action.

An object hangs from a rope. The tension in the string is 8.86 N when it is in the air and 7.84 N when it is submerged in water. What is the density of the object?



8684.2 kg/m3


String tension due to weight = 8.86 N

String tension when immersed in water = 7.84

upthrust = 8.86 - 7.84 = 1.02 N = displacement x acceleration due to gravity

Mass of water discharged = 1.02 / 9.81 = 0.104 kg

Water density = mass of water / volume of water

Make volume the subject of a formula.

Discharged water = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³

volume of object = volume of water displaced

Object density = object mass / object volume = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³

A 71 kg skier climbs a slope on a motorized cable. The acceleration due to gravity is 9.8 m/s2. How much work 2.58 m/s is required to pull your 62 mu up a 31° incline (assuming no friction) at a constant speed? the unit is kJ



The work done will be 22216.894 J


We have given the mass of the skier m = 71 kg

gravitational acceleration

the slope is

distance d = 62 m

that is, the vertical distance

We know that the work done by

A puck with a moment of inertia of 5.0 kg. m2 and a radius of 0.25 m rotates about a fixed, frictionless axis perpendicular to the disk and passing through the center of the disk. A force of 8.0 N is applied tangentially to the edge. If the disk starts to freeze,
So, after you have made half a revolution, your angular velocity is:
(1) 0,57 rad/s
(2) 0,64 rad/s
(3) 1,6 rad/s
(4) 3,2 rad/s


Answer: 1.6 rad/s



moment of inertia of the disk

The radius of the disk



to use

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