# Linear Motion Homework Problem

• _Pistolkisses
In summary, the problem involves adjusting a garden hose nozzle to create a hard stream of water and then pointing it vertically upward at a height of 1.5m above the ground. When the nozzle is quickly moved away from the vertical, the sound of the water striking the ground is heard for 2.0 seconds. The equations used are d = v2t-(0.5)at^2 and v1^2= (v2^2)-2ad, with v2=0, a=-9.8, and t=2.0. The calculated water speed is 18.84m/s, but this is incorrect due to using the wrong distance and not accounting for the signs of the other quantities.

## Homework Statement

You adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5m above the ground. When you quickly move the nozzle away from the vertical, you hear the water striking the ground next to you for 2.0 s. What is the water speed as it leaves the nozzle?

## Homework Equations

I'm guessing, d = v2t-(0.5)at^2

## The Attempt at a Solution

I subbed in v2=0, a = -9.8, and t= 2.0 in the first equation to get distance.

And then subtracted 1.5 from the distance and used the new distance in the second equation (v1^2= v2^2 -2ad) to find velocity and came up with 18.84m/s.

It's not right because I had no idea which distance to use or how to find the right distance because the time value of 2.0 seconds is confusing me (the wording, anyways).

Should distance be -1.5m instead? :S

_Pistolkisses said:
Should distance be -1.5m instead? :S

That's right. The water which leaves the nozzle with a speed of vi upward takes 2 secs to fall -1.5 m from the starting point, if up is taken as positive. Be careful about the signs of the other quantities.

I would suggest using the equation d = v1t + (0.5)at^2 to calculate the distance, where d is the distance, v1 is the initial velocity (in this case, 0 m/s), t is the time (2.0 s), and a is the acceleration due to gravity (-9.8 m/s^2). This will give you the distance from the nozzle to the ground, which should be 5.8 m.

Then, using this distance, you can use the equation v2^2 = v1^2 + 2ad to solve for the final velocity (v2). Substituting in v1 = 0 m/s, a = -9.8 m/s^2, and d = 5.8 m, you would get a final velocity of 18.84 m/s, which is the same result you got.

To answer the question about which distance to use, it would be the distance from the nozzle to the ground, which would be 5.8 m. This distance represents the height at which the water leaves the nozzle and begins to fall due to gravity. Therefore, it is the distance at which you would use to calculate the final velocity.

I hope this explanation helps and clarifies any confusion. Keep up the good work with your homework problems!

## 1. What is linear motion?

Linear motion, also known as rectilinear motion, refers to the movement of an object in a straight line at a constant speed or velocity.

## 2. How do I solve a linear motion homework problem?

To solve a linear motion homework problem, you will need to identify the given variables such as initial velocity, final velocity, time, and acceleration. Then, you can use the appropriate formula (such as d = vt, v = u + at, or v^2 = u^2 + 2as) to calculate the unknown variable.

## 3. What is the difference between speed and velocity in linear motion?

Speed refers to the rate at which an object is moving, while velocity refers to the rate of change of an object's position in a specific direction. Speed does not have a direction, while velocity does. In linear motion, speed is constant while velocity can change due to changes in direction or acceleration.

## 4. Can a linear motion problem have multiple solutions?

Yes, a linear motion problem can have multiple solutions. This can happen when there are multiple unknown variables and given information, or when there are different initial conditions for the same problem.

## 5. What are some real-life examples of linear motion?

Some examples of linear motion in everyday life include a car moving on a straight road, a ball rolling down a ramp, a person walking in a straight line, or a train moving along a track.