Saturday

January 31, 2015

January 31, 2015

Posted by **Josh** on Sunday, September 17, 2006 at 12:37pm.

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.

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