Can someone help me???

This should be simple but I am really struggling. can you explain why the answer is 22m/s and 38 degrees??,

Not enough information provided.

You need to post the whole question for a definitive answer. It's pretty certain this is a vector addition problem, but without all of the information it's guesswork to try and answer you.

Not enough information provided.

You need to post the whole question for a definitive answer. It's pretty certain this is a vector addition problem, but without all of the information it's guesswork to try and answer you.

Thanks.

Ignoring the river velocity for a moment, if the new boat heads off at 1.5 m/s at an angle of 65^o to the riverbank, then resolving the horizontal and vertical components produces:

$\mathrm{horizontal \ velocity} = 1.5cos(65) = 0.634 \mathrm{\ m/s}$

$\mathrm{vertical \ velocity} = 1.5sin(65) = 1.36 \mathrm{\ m/s}$

This is in addition to the river velocity of 1.1 m/s

i.e. the boat must actually travel at

1.36 m/s across the river and,

1.1 + 0.634 = 1.734 m/s downstream

Combining these produces the resultant velocity of:

$\sqrt{(1.36^2 + 1.734^2)} = 2.2 \mathrm{\ m/s}$

at an angle of:

$\theta = tan^{-1}\frac{1.36}{1.734} = 38^o$

There is enough information; draw the first vector at 65 degrees and then the next one horizontal to the page (like in the diagram) joining on to it at the correct magnitudes.

Another question I am struggling with. why would you do cos65 when the angle the 25 degrees to the horizontal

We need to resolve the forces acting on the geologist (in a similar way to resolving the velocity components of the last example) using a reference that puts the forces at right angles to each other and in the direction the geologist must travel up the slope.





Since the geologist is on a mountain slope and not on horizontal ground, he can only move up the slope (hypotenuse), so the forces of gravity and friction need to be resolved. There are two resolved component forces acting at right angles: - the normal force perpendicular to the slope and the frictional force parallel to the slope.

Using the trigonometry rules, we can draw the graphical representation of the force magnitudes and directions:




At this point, we need to introduce the trigonometry rules for the angles of similar triangles. Notice where theta occurs (three places but only two shown explicitly in the diagram above).

Notice that the actual force of gravity acts vertically downwards so the magnitude of that force is given by:

$F_g = mg$

Which makes the normal (to the slope) perpendicular component force:

$F_n = F_gCos\theta$

And the frictional parallel (to the slope) component force:

$F_f = F_gSin\theta$

If I'm being honest I am still confused.