projectile motion Watch

Me123456789
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A ball was projected horizontally at a speed of 0.52ms-1 across the top of an inclined board of width 600mm and length 1200mm. It reached the bottom of the board 0.9s later. Calculate
a) distance travelled by the ball across the board (it's 0.468m)
b) its acceleration on the board
c) speed at bottom of the board

Can someone help me with part b and c please? The answer is 3.0 ms-2 for b and 2.7 ms-1 for c.
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Me123456789
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Qwertish
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Was there a diagram for this question?

Is the ball being projected along the slope of the board, or horizontally so that is undergoes free-fall until it hits the board again? From (b), I imagine it's the first one, but I'm not sure.
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Username_valid
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Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.
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anonymouspie227
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(Original post by Username_valid)
Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.
This.
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Me123456789
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(Original post by Qwertish)
Was there a diagram for this question?

Is the ball being projected along the slope of the board, or horizontally so that is undergoes free-fall until it hits the board again? From (b), I imagine it's the first one, but I'm not sure.
Yeah I've attached it.

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Me123456789
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(Original post by Username_valid)
Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.
Thank you.
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Qwertish
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(Original post by Me123456789)
Yeah I've attached it.

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Oh. Haha, a combination of both then :s

You've done part (a), so for part (b):

You must remember acceleration is a vector quantity, so you must do it separately across the board and down the slope of the board. Across the board, the acceleration is zero, but down the board it is not. So:

s=1.2m
u=0 m/s
v
a=?
t=0.9s

So,


s = ut + \dfrac{1}{2}at^2



a = \dfrac{2(s - ut)}{t^2}

Putting the numbers in gives you 2.96m/s2.
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Me123456789
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(Original post by Qwertish)
Oh. Haha, a combination of both then :s

You've done part (a), so for part (b):

You must remember acceleration is a vector quantity, so you must do it separately across the board and down the slope of the board. Across the board, the acceleration is zero, but down the board it is not. So:

s=1.2m
u=0 m/s
v
a=?
t=0.9s

So,


s = ut + \dfrac{1}{2}at^2



a = \dfrac{2(s - ut)}{t^2}

Putting the numbers in gives you 2.96m/s2.
Thank you!
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Me123456789
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(Original post by Qwertish)
Oh. Haha, a combination of both then :s

You've done part (a), so for part (b):

You must remember acceleration is a vector quantity, so you must do it separately across the board and down the slope of the board. Across the board, the acceleration is zero, but down the board it is not. So:

s=1.2m
u=0 m/s
v
a=?
t=0.9s

So,


s = ut + \dfrac{1}{2}at^2



a = \dfrac{2(s - ut)}{t^2}

Putting the numbers in gives you 2.96m/s2.
How did you get s=1.2?

EDIT: Oh sorry it's okay, I know now.
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Qwertish
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(Original post by Me123456789)
How did you get s=1.2?
The length of the board in metres (because I'm resolving down the length of the board, since acceleration is a vector).
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Me123456789
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(Original post by Qwertish)
The length of the board in metres (because I'm resolving down the length of the board, since acceleration is a vector).
Can you help me with part c please?:confused:
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Qwertish
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(Original post by Me123456789)
Can you help me with part c please?:confused:
So, they want the speed (not the velocity), which means you can ignore all the vectors stuff. Write down what you have (you've got the distance, s, initial speed, u, time, t, and you want the final speed, v). Then see which SUVAT equation you need to use .
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Me123456789
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(Original post by Qwertish)
So, they want the speed (not the velocity), which means you can ignore all the vectors stuff. Write down what you have (you've got the distance, s, initial speed, u, time, t, and you want the final speed, v). Then see which SUVAT equation you need to use .
Thanks! :thumbsup:
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Me123456789
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(Original post by Qwertish)
So, they want the speed (not the velocity), which means you can ignore all the vectors stuff. Write down what you have (you've got the distance, s, initial speed, u, time, t, and you want the final speed, v). Then see which SUVAT equation you need to use .
Sorry to keep bothering you, but when your working out anything to do with the vertical is U always 0?
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Me123456789
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(Original post by Username_valid)
Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.
Hey, can you help me with part c of this question please?
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Username_valid
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(Original post by Me123456789)
Thank you.
No problem. To answer your question above, the vertical initial velocity is not always zero but it is in this case because the ball is projected horizontally.

Remember to find speed you have to find the final velocity for both the horizontal and vertical components and find the resultant using Pythagros.
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Qwertish
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(Original post by Me123456789)
Sorry to keep bothering you, but when your working out anything to do with the vertical is U always 0?
The ball is pushed horizontally, so the vertical component of the initial velocity is 0, yes.
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Me123456789
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So would I use

s = 1/2(u+v)t

where t=2.7, s=35.7 (my answer to part b) and u=0 ?
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Me123456789
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(Original post by Username_valid)
No problem. To answer your question above, the vertical initial velocity is not always zero but it is in this case because the ball is projected horizontally.

Remember to find speed you have to find the final velocity for both the horizontal and vertical components and find the resultant using Pythagros.
So would I use

s = 1/2(u+v)t

where t=2.7, s=35.7 (my answer to part b) and u=0 ?
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