# M1 - SUVAT projectiles help?Watch

#1
A ball is thrown at 12 m/s at 60 *degrees* to the horizontal. After 1 s of flight what is its position?

I've tried making SUVAT tables:

x y
S
U
V
A 0
T 1

y
S
U 12
V
A -9.8
T 1

I know you make a righht-angled triangle, with the angle 60 and the hypotenuse the initial vertical velocity (Uy=12). But is the adjacent the horizontal velocity (Ux) and the opposite the vertical displacement (Sy)? And what do you do after that?
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8 years ago
#2
(Original post by rosefan42)
I know you make a righht-angled triangle, with the angle 60 and the hypotenuse the initial vertical velocity (Uy=12). But is the adjacent the horizontal velocity (Ux) and the opposite the vertical displacement (Sy)? And what do you do after that?
You want to find the initial vertical velocity (Uy) and the initial horizontal velocity.

The initial velocity is 12m/s at 60 degrees to the horizontal. Now I'll let you in to a small cheat. Recall that cosine is initially 1 at x=0 and falls down to 0 at x=90. Sine is initially 0 and goes up to 1 at x=90.

If the ball is travelling at 60 degrees to the horizontal then the vertical velocity is going to be greater than the horizontal velocity. Therefore we multiple the initial velocity by the bigger trig value to get the vertical velocity, and by the small trig value to get the horizontal velocity. sin60 > cos60 (as we're nearer x=90 than x=0).

So the vertical velocity is 12sin60, and the horizontal velocity is 12cos60.

You have horizontal velocity, horizontal acceleration, and time. Using that you can work out horizontal position.

Similarly, you have vertical velocity, vertical acceleration, and time. Using that you can work out vertical position.

Now we have horizontal position and vertical position, we're done.
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8 years ago
#3
(Original post by Kolya)
You want to find the initial vertical velocity (Uy) and the initial horizontal velocity.

The initial velocity is 12m/s at 60 degrees to the horizontal. Now I'll let you in to a small cheat. Recall that cosine is initially 1 at x=0 and falls down to 0 at x=90. Sine is initially 0 and goes up to 1 at x=90.

If the ball is travelling at 60 degrees to the horizontal then the vertical velocity is going to be greater than the horizontal velocity. Therefore we multiple the initial velocity by the bigger trig value to get the vertical velocity, and by the small trig value to get the horizontal velocity. sin60 > cos60 (as we're nearer x=90 than x=0).

So the vertical velocity is 12sin60, and the horizontal velocity is 12cos60.

You have horizontal velocity, horizontal acceleration, and time. Using that you can work out horizontal position.

Similarly, you have vertical velocity, vertical acceleration, and time. Using that you can work out vertical position.

Now we have horizontal position and vertical position, we're done.
Where does come from?
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8 years ago
#4
(Original post by steve2005)
Where does come from?
Huh? x is just a variable we we're using to compare cos(x) and sin(x). If x is close to 90, sin(x) is greater than cos x. If x is close to 0, cosx is greater than sinx.
Last edited by Kolya; 8 years ago
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8 years ago
#5
(Original post by Kolya)
Huh? x is just a variable we we're using to compare cos(x) and sin(x). If x is close to 90, sin(x) is greater than cos x. If x is close to 0, cosx is greater than sinx.
i j evaluate and you have the position after 1 second of flight.
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8 years ago
#6
(Original post by steve2005)
i j evaluate and you have the position after 1 second of flight.
It is (rightly) preferred if you tell someone looking for help how to go about solving their problem, rather than simply doing it for them (especially without any explanation).
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8 years ago
#7
(Original post by Kolya)
It is (rightly) preferred if you tell someone looking for help how to go about solving their problem, rather than simply doing it for them (especially without any explanation).
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8 years ago
#8
horizontally isn't it just s =vt so it's 12cos60*1
s = 6m
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8 years ago
#9
(Original post by steve2005)
I'm confused. Which part was misleading?
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8 years ago
#10
(Original post by Kolya)
I'm confused. Which part was misleading?
The introduction of is confusing to me, of course I know that x is a variable BUT where does the 90 come from? i know how to solve the problem so I think it will also be confusing to the OP.
Last edited by steve2005; 8 years ago
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8 years ago
#11
I don't get it, it's thrown to the horizontal, why bother looking at the vertical position then?
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8 years ago
#12
(Original post by boromir9111)
I don't get it, it's thrown to the horizontal, why bother looking at the vertical position then?
It's thrown UP and OUT if you get what I mean.
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8 years ago
#13
(Original post by steve2005)
It's thrown UP and OUT if you get what I mean.
Okay i have a question of my own that i remembered...... it involves a tennis a ball that is thrown to the horizontal as well..... i can't really explain it without drawing it but i will try to explain it..... basically it asks you to measure its initial velocity and show that it's about 18ms^-1

Information we are given that the ball is 3.0m off from the ground, horizontal distance it travels is 11.0m and the net is 0.9m and from this we need to measure u to about 18ms^-1

For this do you look at horizontal and vertical components as well????

edit - cause i'm pretty sure you need "t" in order to get "u" and without "t" you can't get "u"
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8 years ago
#14
can anyone answer my question? i know it's a little vague but try to your best to just imagine it please.
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8 years ago
#15
(Original post by boromir9111)
can anyone answer my question? i know it's a little vague but try to your best to just imagine it please.
Where does the net come into the problem. I think we need a diagram or better explanation/description.

Where does the question come from?
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8 years ago
#16
(Original post by steve2005)
Where does the net come into the problem. I think we need a diagram or better explanation/description.

Where does the question come from?
I have no idea but the way i did it which makes me no sense and yet i get the answer strangely..... not sure how to draw on here but the way i described is literally how it was shown to me, a vertical height of of 3.0m from where the ball is off the ground..... then it shows a horizontal distance of 11.0m and a height of the net which is 0.9m.

edit - came in the physics exam i sat on February 3rd as the first question.
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8 years ago
#17
(Original post by boromir9111)
I have no idea but the way i did it which makes me no sense and yet i get the answer strangely..... not sure how to draw on here but the way i described is literally how it was shown to me, a vertical height of of 3.0m from where the ball is off the ground..... then it shows a horizontal distance of 11.0m and a height of the net which is 0.9m.

edit - came in the physics exam i sat on February 3rd as the first question.
How far from the initial position of the ball is the net? Does the ball just clear the net? Can't you post a diagram?
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8 years ago
#18
(Original post by steve2005)
How far from the initial position of the ball is the net? Does the ball just clear the net? Can't you post a diagram?
I don't remember much about the content cause that wasn't necessary, all i remember is the numbers that were needed but i think the horizontal is distance from the ball to the net which is, 11.0m. No, i installed the new windows on my laptop and ain't installed windows office yet, sorry mate.

edit - don't need windows office i just realized lol, did it on paint instead.
Last edited by boromir9111; 8 years ago
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8 years ago
#19

That's the best i can do really and of course my ball interpretation is a bit off cause there was a racket there as well, etc.
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