Kinematics and Gravity Problem

[i_m_mimi]

Homework Statement

7. A fireman is standing on top of a building 20.0m high. He finds that if he holds the hose so that water issues from it horizontally at 12.0m/s, the water will hit a burning wall of an adjacent building, at a height of 15.0m above the ground. What is the horizontal distance from the fireman to the burning wall? (Already solved)

8. The fireman in Exercise 7 wants the water from the same hose to reach the burning wall at the same level above the ground as he is standing. At what angle must he aim the hose relative to the horizontal?

Homework Equations

vf =vo +2at
d=vot + 1/2at²
vf² = vo² +2ad
g =9.8m/s²

I already solved exercise 7 and I found the distance between the buildings is 12.1m.

The correct answer to exercise 8 is 27.7 degrees.

The Attempt at a Solution

First I tried solving exercise 8 as if the horizontal component of the velocity is 12.0, but that didn't work out.
Then I tried solving it with the angled velocity as 12.0m, then finding horizontal and vertical component velocities of that, but I'm having trouble at this point on how to do that. Then after I would use trig to find the angle.

v = d/t
vx = 12.1m/t
t = 12.1m /vx
vx squared + vy squared = 12.0m squared
d=vot + 1/2at²
I tried combining 2 equations but that the answer that I get doesn't work either.

:/

Thanks for any help.

 

[cepheid]

You want to find the angle such that the parabolic path of the water goes up, stops and comes back down to the same level all within the span of the 12.1 m horizontal distance between the buildings. So here are the steps I would take to solve this problem:

1. Figure out the time needed to reach the max height (solve for it in terms of the other unknown variables). In other words, this is the time needed for the vertical velocity to reach zero (starting from its initial value).

2. By symmetry, the time needed to come back down to the same level from the max height is the same as the time needed to go up to it, so the total time is just twice the value you found in 1.

3. Figure out the horizontal component of the velocity based on the horizontal distance and the total travel time found in 2.

4. Now that you know the horizontal velocity (probably in terms of the vertical velocity), and you know the total velocity, you can find both components. From the two components, you can find the launch angle.

 

[PeterO]

Homework Statement

7. A fireman is standing on top of a building 20.0m high. He finds that if he holds the hose so that water issues from it horizontally at 12.0m/s, the water will hit a burning wall of an adjacent building, at a height of 15.0m above the ground. What is the horizontal distance from the fireman to the burning wall? (Already solved)

8. The fireman in Exercise 7 wants the water from the same hose to reach the burning wall at the same level above the ground as he is standing. At what angle must he aim the hose relative to the horizontal?

Homework Equations

vf =vo +2at
d=vot + 1/2at²
vf² = vo² +2ad
g =9.8m/s²

I already solved exercise 7 and I found the distance between the buildings is 12.1m.

The correct answer to exercise 8 is 27.7 degrees.

The Attempt at a Solution

First I tried solving exercise 8 as if the horizontal component of the velocity is 12.0, but that didn't work out.
Then I tried solving it with the angled velocity as 12.0m, then finding horizontal and vertical component velocities of that, but I'm having trouble at this point on how to do that. Then after I would use trig to find the angle.

v = d/t
vx = 12.1m/t
t = 12.1m /vx
vx squared + vy squared = 12.0m squared
d=vot + 1/2at²
I tried combining 2 equations but that the answer that I get doesn't work either.

:/

Thanks for any help.

If the water is to hit the building at the same height, the problem is the same as the standard "projectile on level ground" examples.
Analysis of vertical and horizontal velocity components lead to a simple range formula which depend only on initial velocity, gravity and the angle [se http://en.wikipedia.org/wiki/Range_of_a_projectile]

That will give an answer pretty easily.

NOTE: any chance this problem was supposed to be solved using a g value of 10 m.s^-2, as that would very quickly yield a building separation of 12m without any decimal arithmetic. We often use g=10 in introductory Physics so that messy arithmetic does not interfere with Physics understanding. [if only those early French scientists had made the metre a tiny bit shorter, g would have had a value of 10 !]