- #1

Toranc3

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## Homework Statement

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00s to reach a building 45.0 away. You can ignore air resistance; assume that the end of the hose is at ground level.

a. Find the angle of elevation

b. Find the speed and acceleration of the water at the highest point in its trajectory

c.how high above the graound does the water strike the building and how fast is it moving just before it hits the building?

## Homework Equations

x-xo=vox*t

vx=vox

y=yo +voy*t -1/2 *g *t^(2)

vy=voy-g*t

y-yo=(voy+vy)/2 *t

vy^(2)=voy^(2) -2g(y-yo)

## The Attempt at a Solution

a= 53.1 degrees

x=xo+vox*t

45.0m=25.0m/s*cos(theta)*3.00s

Theta=53.1 degrees

b=15 m/s and g=9.8m/s^(2)

Vox=25.0m/s*cos(53.1)= 15m/s

Aceleration is the same vertically.

c=15.9m

y=yo+voy*t -1/2*g*t^(2)

y= 20m.s*3.00s-1/2*g*9.00s^(2)

y=15.8m

d=17.7m/s

vy^(2)=voy^(2)-2g(y-yo)

vy^(2)=(20m.s)^(2)-19.6m/s^(2)*15.9m

vy=9.4m/s

therefore Vf=17.7m/s by v=sqrt(vx^(2)+(vy)^(2))

vx=15.0m/s

My question is is that when I try to check my answer with other equations they do not come out the same. Why is that?

For example:

I want to get my heigth at the highest point

vy^(2)=voy^(2)-2g(y-yo)

0=(20m/s))^(2)-19.6*y i get y=20m

y-yo=(voy+vy)/2 * t

y-0=(20m/s+0)/2*3.00s here i get 30m

How come these equations won't work to give me the actual height of 15.9m?

## Homework Statement

## Homework Equations

## The Attempt at a Solution

## Homework Statement

## Homework Equations

## The Attempt at a Solution

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