Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 21.0 m/s as it leaves the hose nozzle. 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 3.00 s to reach a building 35.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.

Find the angle of elevation of the hose.
Got it.... 56.3 degrees

Find the speed of the water at the highest point in its trajectory.
Don't know...

Find the acceleration of the water at the highest point in its trajectory.
Got it.... 9.8 m/s^2

How high above the ground does the water strike the building?
Got it..... 8.28 m

How fast is it moving just before it hits the building?
Yeah...don't know... :)

As I recall, you asked this multipart question before. The method of doing all parts was described in my previous answer.

Here again you should try to set up a position and velocity vector. A projectile motion means the path of a parabola or second degree equation.
The position vector is given by
(1) (x,y)=(cos(p)*v*t,(-1/2)g*t^2+sin(p)*v*t)
The velocity vector is given by
(2) (v_x,v_y)=(cos(p)*v,-g*t+sin(p)*v)]
Here v=2.0m/s, distance traveled = 35.0m and the time taken is 3.0s
You should be able to determine the angle using the x component in (1) and the given values.
Speed is the magnitude of the velocity. At the highest point the v_y component is 0, so the speed will be the absolute value of the v_x component.
The acceleration component is always constant. I think it should be negative considering how the problem was set up.
For the last part "How fast is it moving just before it hits the building?" I think the question wants the speed of the water, not just one of the components of velocity.
Speed=magnitude of velocity = sqrt((v_x)^2 + (v_x)^2)
Use t=3, since the water's in the air for that long, and solve speed using (2) above.

Hope this helps.

Ah, I see. Well, let's calculate the speed of the water at the highest point in its trajectory using the given values. We know that the water reaches the highest point when its vertical velocity component is zero. Therefore, we can use the equation v_y = v*sin(theta) - g*t, where v is the initial speed of the water, theta is the angle of elevation, g is the acceleration due to gravity, and t is the time taken for the water to reach the building.

Since v_y = 0 at the highest point, we can rearrange the equation to solve for v*sin(theta).

0 = v*sin(theta) - g*t

v*sin(theta) = g*t

Now, let's plug in the given values. We know that g is approximately 9.8 m/s^2 and t is 3.00 s.

v*sin(theta) = 9.8 m/s^2 * 3.00 s

v*sin(theta) = 29.4 m/s

Now, to find the speed of the water at the highest point, we just need to divide both sides of the equation by sin(theta).

v = 29.4 m/s / sin(theta)

Since we know that the angle of elevation is 56.3 degrees, we can substitute that value into the equation.

v = 29.4 m/s / sin(56.3 degrees)

Using a calculator, we find that v is approximately 39.6 m/s.

So, the speed of the water at the highest point in its trajectory is approximately 39.6 m/s.

I hope that answers your question!

To find the speed of the water at the highest point of its trajectory, we can analyze the velocity vector at that point. The y-component of the velocity at the highest point is 0 m/s, so we can set that equal to 0 in equation (2):

0 = -gt + sin(p) * v

Solving for sin(p):

sin(p) = gt/v

Plugging in the given values of g = 9.8 m/s^2 and v = 21.0 m/s, we can solve for sin(p):

sin(p) = (9.8 m/s^2 * 3.00 s) / 21.0 m/s
sin(p) = 1.40

Since the angle of elevation cannot be greater than 90 degrees, we know that sin(p) must be less than or equal to 1. So, there is no solution for this equation. This means that there is no highest point in the water's trajectory and it never reaches a vertical peak. Therefore, the water does not have a specific speed at the highest point of its trajectory.

For the last question, we can find the speed of the water just before it hits the building. As mentioned earlier, the speed is the magnitude of the velocity vector. Using equation (2), we can find the x-component and y-component of the velocity just before it hits the building:

v_x = cos(p) * v = cos(56.3 degrees) * 21.0 m/s
v_x ≈ 10.5 m/s

v_y = -g * t + sin(p) * v = -9.8 m/s^2 * 3.00 s + sin(56.3 degrees) * 21.0 m/s
v_y ≈ -29.4 m/s

Now, we can calculate the speed:

speed = √(v_x^2 + v_y^2) = √((10.5 m/s)^2 + (-29.4 m/s)^2)
speed ≈ 31.2 m/s

Therefore, the water is moving at a speed of approximately 31.2 m/s just before it hits the building.

To find the speed of the water at the highest point in its trajectory, we need to consider the velocity vector (v_x, v_y) at that point. Since the water is at its highest point, the vertical component of velocity, v_y, is equal to 0. Therefore, we can use the equation:

v = sqrt((v_x)^2 + (v_y)^2)

where v is the magnitude of velocity. Substituting v_y = 0, we have:

v = sqrt((v_x)^2 + 0^2)

v = |v_x|

So, the speed of the water at the highest point in its trajectory is equal to the absolute value of the horizontal component of velocity, v_x. To find v_x, we can use equation (2) mentioned earlier:

v_x = cos(p) * v,

where p is the angle of elevation of the hose. From the previous part of the question, we found that the angle of elevation is 56.3 degrees. Substituting this value and the given speed v = 21.0 m/s, we have:

v_x = cos(56.3°) * 21.0 m/s

By calculating this, we get:

v_x ≈ 11.0 m/s

Thus, the speed of the water at the highest point in its trajectory is approximately 11.0 m/s.

Moving on to the next part, to find the acceleration of the water at the highest point in its trajectory, we can use the equation for acceleration:

a = -g

where g is the acceleration due to gravity, approximately 9.8 m/s^2. Therefore, the acceleration of the water at the highest point is 9.8 m/s^2 in the downward direction.

Finally, to determine how high above the ground the water strikes the building, we need to find the y-component of the position vector (x, y) at the time it reaches the building. Using equation (1) mentioned earlier:

y = (-1/2)g*t^2 + sin(p)*v*t

Substituting g = 9.8 m/s^2, t = 3.00 s, p = 56.3 degrees, and v = 21.0 m/s, we have:

y = (-1/2)(9.8 m/s^2)(3.00 s)^2 + sin(56.3°)(21.0 m/s)(3.00 s)

By calculating this, we get:

y ≈ 8.28 m

Therefore, the water strikes the building approximately 8.28 m above the ground.

For the last part of the question, "How fast is it moving just before it hits the building?", we can find the speed by calculating the magnitude of the velocity vector (v_x, v_y) at that point. Using the equation for speed mentioned earlier:

speed = sqrt((v_x)^2 + (v_y)^2)

Substituting v_x ≈ 11.0 m/s (from earlier calculations) and v_y = -g*t (since the water is moving downward), we have:

speed = sqrt((11.0 m/s)^2 + (-9.8 m/s^2)(3.00 s))^2

By calculating this, we can find the speed of the water just before it hits the building.