Physics Resolving Vectors Help !!!

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Lyrapettigrew
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#1
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#1
Hello sorry to ask but I am truly stumbling about how to answer these questions:

a) Firstly, calculate the horizontal and vertical components the initial velocity of a ball thrown into the air, projected at sixty degrees to the horizontal.
b. A man is exerting a force of 100N on a lawn mower at 30° to the horizontal. Find the horizontal and vertical components of the force.
c. A block of 50N weight is resting on a plank inclined at 40o to the horizontal. Calculate the components of its weight parallel to the plank and perpendicular to the plank.

I do not understand how to solve the part a of the question considering the direction but not the magnitude of the resultant velocity is given.

For b) I understand that forces are vectors and that one can use trigonometry to determine the resultant vector and direction. In this case, picturing a right angled triangle where the hypotenuse is equal to the resultant force (100N) at an angle of 30 degrees one could calculate the horizontal component by:
cos 30 = Adj/Hyp
cos 30 = Adj/100
Adj= 100 * cos 30
Adj = 86.6 N to 3.s.f.

To calculate the vertical component:
sin 30 = Opp/Hyp
sin 30=Opp/100
Opp=100 * sin 30
Opp=50 N

For c) The horizontal component could be found by:
cos 40 = Adj/Hyp
cos 40 = Adj/50
Adj= 50 * cos 40
Adj= 38.3 N (3.s.f)
The vertical component:
sin 40 = Opp/Hyp
sin 40=Opp/50
Opp=50 * sin 40
Opp=32.1 N (3.s.f)

I am really unsure if any of my approach is correct and would really appreciate a further explanation of how to solve this type of question (and how to begin to resolve part a).

Thank you for any and all help, I offer my sincerest gratitude 😁
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Yodalam
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#2
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#2
Part A is the most simple-
The horizontal component of the velocity is simply: Ucos60
The vertical component of the velocity is: Ucos60
U being the initial velocity
These can be worked out using Cos, sin equations.

Part B is correct.

Part C is also correct

You understand the concept
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Lyrapettigrew
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#3
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#3
(Original post by Yodalam)
Part A is the most simple-
The horizontal component of the velocity is simply: Ucos60
The vertical component of the velocity is: Ucos60
U being the initial velocity
These can be worked out using Cos, sin equations.

Part B is correct.

Part C is also correct

You understand the concept
Fabulous, thank you so much for your reply. I was thinking of solving part a that way initially but thought that it was a little unfinished and that perhaps I was missing something. So are the final answers for the horizontal component u cos 60o and for the vertical u sin 60o?
Thank you for your explanation and confirmation. 😁
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Yodalam
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#4
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#4
Yes the answer for A are correct
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Lyrapettigrew
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#5
Report Thread starter 2 years ago
#5
(Original post by Yodalam)
Yes the answer for A are correct
Thank you very much for all of your help, have great day ✌️
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