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September 30, 2014

September 30, 2014

Posted by **JC** on Monday, September 18, 2006 at 5:30pm.

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?

don't know!

Are you by chance referring to the help I've given on this J.C.? I'll try again and walk through the steps if you want - I won't just give the answer though.

I would be interested to know if your course is calculus based or not. If it's, not that's ok, it just helps me to know what can be used here. You also need to be aware that this problem is the beginning of solving elementary mechanics using vectors, so I'm supposing you've seen these already. Try to be as specific as you can on the difficulties too, that helps in explaining the answer.

I see there are typos in my first post and I probably could've been a little clearer in the explanation. When we first study motion we study linear motion. After that we study 2-dimensionsal or motion in the plane. I'm not sure if you've studied what projectile motion looks like or not. Basically when an object is projected near the surface of the earth it's path looks like a parabola. A parabola is the best first approximation we can use for projectiles. All parabolas can be defined by the equation ax^2+bx+c or a 2nd degree equation. It is important that you know that the motion vetically is independent of the motion horizontally. When we study 2-d motion we study the component-wise and treat each component as simple linear motion in that direction.

These are the formulas I gave in my previous post on this. The position vector is given by

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

When the object is thrown at an angle p, it's horizontal velocity component is given by cos(p)*v and it's vertical velocity component is given by sin(p)*v.

The velocity vector is given by

(2) (v_x,v_y)=(cos(p)*v,-g*t+sin(p)*v)]

This is the vector that gives the velocity of the particle at any time.

Whatever text you're using should have something similar to the formulas used here.

Here v=21.0m/s, distance traveled = 35.0m and the time is 3.0s

You should be able to determine the angle using the x component in (1) and the given values. The x component in (1) is cos(p)*v*t.

The water travels 35.0m in 3.0s, so 35.0m=cos(p)*21.0m/s *3s so

cos(p)=35/63. arccos(35/63)=56.3deg which you got.

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 velocity in the x or horizontal direction is constant and = cos(p)*v, so speed =|cos(p)*v| = .555 * 21m/s = 11.7m/s

The acceleration component is always constant. You might be correct on using just 9.8m/s^2, check for a worked example in your text to be sure.

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_y)^2)

Check my work

Speed= sqrt(cos(p)*v)^2 + (-g*t+sin(p)*v))^2)

Using p=56.3, v=21.0m/s, g=9.8m/s^2

Speed=sqrt(cos(p)*v)^2 + (-g*t+sin(p)*v))^2)

Speed=sqrt((cos(56.3)*21.0m/s)^2 + (-9.8m/s^2*3+sin(56.3)*21.0m/s)^2)

I'll let you do the calculations.

Please repost and ask questions if anything's unclear.

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