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keweezz
Aug27-08, 09:25 PM
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 14 m with the same initial speed of 13 m/s, but in different directions. You through your snowball downward, at 40° below the horizontal; your friend throws her snowball upward, at 40° above the horizontal. What is the speed of each ball when it is 5.0 m above the ground? (Neglect air resistance.)


m/s (your snowball)
m/s (your friend's snowball)

15.7 m/s is what I keep getting... I did 13m/s sin 30 = 8.3 and.. i used that for Vf^2 = vi^2 + 2ax and used -9 for x and -9.81 for a...

well 15.7 is not right... any thought?

chislam
Aug27-08, 09:36 PM
Why did you use -9 for x? First, find how long each snowball will be in the air.

granpa
Aug27-08, 09:41 PM
sin 30 or 40?

LowlyPion
Aug27-08, 09:44 PM
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 14 m with the same initial speed of 13 m/s, but in different directions. You through your snowball downward, at 40° below the horizontal; your friend throws her snowball upward, at 40° above the horizontal. What is the speed of each ball when it is 5.0 m above the ground? (Neglect air resistance.)


m/s (your snowball)
m/s (your friend's snowball)

15.7 m/s is what I keep getting... I did 13m/s sin 30 = 8.3 and.. i used that for Vf^2 = vi^2 + 2ax and used -9 for x and -9.81 for a...

well 15.7 is not right... any thought?

Remember velocity is composed of 2 components. What is the horizontal (x) component of velocity. That should stay the same with no air resistance. Then you have the vertical component to deal with. That is where you need to take into account the effect of gravity.

keweezz
Aug27-08, 10:20 PM
i used -9 because that was the difference between 5-14? and its sin 40 :O

so i find t and use Vi+at=Vf?

i find t using x=Vit+1/2a(t)^2?

LowlyPion
Aug27-08, 10:29 PM
i used -9 because that was the difference between 5-14? and its sin 40 :O

so i find t and use Vi+at=Vf?

i find t using x=Vit+1/2a(t)^2?

Yes that will give you the vertical velocity component. But the distance you should use is +9 as that is in the direction your velocity and acceleration is.

Then you need to combine that with the horizontal component. (Velocity is a vector. You add the two vectors (x,y) together.)

Edit: Btw there is nothing wrong with using the Vf2 = Vo2 +2aΔx you started out with.