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[Worksheet] Horizontal/Vertical components + other

  • Thread starter Torald
  • Start date
2
0
Hey there.

Was given this sheet a while ago. Went over it in class a week ago, completely forgot everything about it today (This is a fresh sheet, lost the one I did.). Have an NAB tomorrow, and couldn't find my physics teacher today to get this worked out.

So, the problem is that I have no clue with what to do. I got some basic info down, but that is it.

Course is in Higher Physics.


Thanks in advance.

Edit : FORGOT THE WORKSHEET!

Phys_001.jpg
 
33,373
9,098
I don't understand what you wrote there.

For (a), you should consider a triangle, where one angle is 50° and the other one is 90° ("standing" on the ground). This will give a relation between horizontal speed and total speed, as well as a relation between vertical and total speed via trigonometry.

For b: Does the horizontal velocity change? If not, how can you calculate the time until the ball reaches the net?

c: What is the vertical position of the ball after that time?
 
2
0
Thanks for the reply.

U = Initial Velocity (6.0 m/s)

Hv = Horizontal Component (2.0m)

Vv = Vertical Component (0.9m)

Focusing on question a):

[STRIKE]Trigonometry...Egh. Hate it.

Since the Hv is perpendicular (90 degrees), to calculate the components, it would involve cos(x), right?
Don't know what do with that, though. For the Hv, would I do (As in, punt into a calculator) "6.0cos(90)"? Or "2.0cos(90)"? Or....?

As I said, I have not much of a clue as to what to do here.

Your help is most appreciated.

Edit: Alright. I think I figured out how to calculate the components.

Vv = 6sin(50) = 3.9

Hv = 6cos(50) = 4.5

But shouldn't they add up to 6? 3.9 + 4.5 = 8.4

Gah. Confuzzled.[/STRIKE]

Got it!

The Hv = Vcos(Θ)
= 6cos(50)
= 3.9

Trick to remember : HC - Horizontal Cos


The Vv = Vsine(Θ)
= 6sine(50)
= 4.5

Trick to remember : VS - Vertical Sine



For

D = 2.0
V = 6.0

So, use d=v/t |With rearranging| t = d/v

2.0/6.0 = 0.3



Although, I don't know what to do for [c]. Help would be good. I assume it involves the Vv, Velocity, time.
 
Last edited:
30
0
For c) so at 3 s it reaches the net. How far will it travel vertically in 3s with initial velocity 6sin30? If this is greater than 0.9 then it makes it over the net.
 
33,373
9,098
The Vv = Vsine(Θ)
= 6sine(50)
= 4.5
You should be careful with rounding (4.596...)

For

D = 2.0
V = 6.0

So, use d=v/t |With rearranging| t = d/v

2.0/6.0 = 0.3

No, it is a horizontal distance, and the horizontal speed is not 6m/s.


@ofeyrpf: Don't forget gravity.
 

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