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**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

A fireman has a hose and is standing on top of a 20 m high building. His hose shoots water at 12 m/s and he wishes to hit the top of another 20 m hig building 21 m away. what angle should he aim his hose at?

I calculated that in the x direction, the velocities are both 12cosθ, the acceleration is zero, distance is 21.0 and te time is unknown. In the y direction, te initial velocity is 12sinθ and the final is -12sinθ, the acceleration is -9.8, the distance is zero, and the time is also unknown.

I then set up a system:

x: t=21/12cosθ

y: t=24sinθ/9.8

0=21/12cosθ-24sinθ/9.8

(21/12)secθ=(24/9.8)sinθ

(21/12)/(24/9.8)secθ=sinθ

sinθ/secθ=205.8/288

sinθcosθ=205.8/288

I'm stuck now...

*Last edited by fusilli_jerry89 (2006-10-23 07:52:29)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

from what youve posted, there is no answer

because it would result in taking the inverse sine of a number round about the size of 14 which is outside its range

*Last edited by luca-deltodesco (2006-10-23 10:29:40)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

fusilli_jerry89 wrote:

A fireman has a hose and is standing on top of a 20 m high building. His hose shoots water at 12 m/s and he wishes to hit the top of another 20 m hig building 21 m away. what angle should he aim his hose at?

I calculated that in the x direction, the velocities are both 12cosθ, the acceleration is zero, distance is 21.0 and te time is unknown. In the y direction, te initial velocity is 12sinθ and the final is -12sinθ, the acceleration is -9.8, the distance is zero, and the time is also unknown.

if you take the angle to be from the x axis counter clockwise

you require that s_x = 21m. so you can calculate t for theta by

then you also require that for this value of theta, s_y will be 0, since buildings are at same height, i.e.

but again, its outside of the range?

anyone tell me where im going wrong?

*Last edited by luca-deltodesco (2006-10-23 10:39:24)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You're not going wrong, it's just that the hose is too weak. The optimum angle of projection is 45° (when the start and finish are at the same height), and that has a range of v²/g, where v is the initial velocity of the projection and v is the gravitational force.

So in this situation, the maximum range of the water from the hose would be 144/9.8, which is around 14.7m, falling considerably short of the 21m target.

Why did the vector cross the road?

It wanted to be normal.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

mathsyperson wrote:

You're not going wrong, it's just that the hose is too weak.

good, because i went over that twice, and i was getting worried.

The Beginning Of All Things To End.

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