+0  
 
0
534
7
avatar+1832 

 

 

 

 

 

 

 

 

xvxvxv  Oct 25, 2014

Best Answer 

 #1
avatar+27228 
+10

Plane

 

A. The angle between south and southwest is 45°. 

The x-component (east) of velocity is -16*cos(45°):  vx = -16/√2 m/s

The y-component (north) of velocity is -(16*sin(45°) + 45):  vy = -(16/√2 + 45) m/s

 

B. Magnitude:  vP/E = √(vx2 + vy2)

 

C.  Direction:  θ = tan-1(vy/vx)  But this will give a result in the first quadrant (i.e. an angle between 0° and 90°).  You will need to convert this to an angle in the third quadrant (i.e. between 180° and 270°).

Alan  Oct 25, 2014
 #1
avatar+27228 
+10
Best Answer

Plane

 

A. The angle between south and southwest is 45°. 

The x-component (east) of velocity is -16*cos(45°):  vx = -16/√2 m/s

The y-component (north) of velocity is -(16*sin(45°) + 45):  vy = -(16/√2 + 45) m/s

 

B. Magnitude:  vP/E = √(vx2 + vy2)

 

C.  Direction:  θ = tan-1(vy/vx)  But this will give a result in the first quadrant (i.e. an angle between 0° and 90°).  You will need to convert this to an angle in the third quadrant (i.e. between 180° and 270°).

Alan  Oct 25, 2014
 #2
avatar+1832 
0

yes thats good .. thank you alan 

 

what about the second one 

xvxvxv  Oct 25, 2014
 #3
avatar+27228 
+5

Baseball

 

A. Horizontal component of velocity is vx = v*cos(43°)

    Vertical component of velocity is vy = v*sin(43°)

 

Horizontal velocity is constant so time to hit ground is t = 188/vx

In this time vertical distance travelled is -0.9m so:

-0.9 = vy*t - (1/2)*g*t2 where g = 9.8m/s2

 

so: -0.9 = vy*188/vx - (1/2)*9.8*(188/vx)2

 

-0.9 = 188*tan(43°) - (1/2)*9.8*(188/cos(43°))2/v2

 

v2 = (1/2)*9.8*(188/cos(43°))2/(188*tan(43°) + 0.9)  m2/s2

 

$${\mathtt{v}} = {\sqrt{{\frac{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{188}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{43}}^\circ\right)}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{188}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{43}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.9}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{42.865\: \!673\: \!803\: \!496\: \!16}}$$

 

v ≈ 43 m/s

Alan  Oct 25, 2014
 #4
avatar+94105 
0

Alan, I am really confused.  Nothing new there.  I've only looked at the first one.

Does the air speed indicator show the actual speed, or the speed if there was no crosswind?

Could you draw the picture please.  I don't get it.    

Melody  Oct 25, 2014
 #5
avatar+27228 
+5

Does this help?

vectors

.

Alan  Oct 25, 2014
 #6
avatar+94105 
0

Yes thanks Alan.  Now it is very clear.  

Melody  Oct 25, 2014
 #7
avatar+1832 
0

Ok thank you alan its very nice =D 

xvxvxv  Oct 25, 2014

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