# Projectile Motion Question

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#1
hey guys, struggling to answer this question any help would be appreciated

A bullet is fired horizontally at a speed on 200ms-1 at a target which is 100m away. Ignoring A.R., calculate how far the bullet has fallen when it hits the target. Also find the angle the bullet makes with the horizontal at this instance

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1 month ago
#2
(Original post by anon8rfy7u)
hey guys, struggling to answer this question any help would be appreciated

A bullet is fired horizontally at a speed on 200ms-1 at a target which is 100m away. Ignoring A.R., calculate how far the bullet has fallen when it hits the target. Also find the angle the bullet makes with the horizontal at this instance

What is your problem? I presume you're analysing horizontal and vertical motion seperately?
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1 month ago
#3
(Original post by anon8rfy7u)
hey guys, struggling to answer this question any help would be appreciated

A bullet is fired horizontally at a speed on 200ms-1 at a target which is 100m away. Ignoring A.R., calculate how far the bullet has fallen when it hits the target. Also find the angle the bullet makes with the horizontal at this instance

As mqb2766 has suggested, you need to split the motion into its horizontal and vertical components.

If the bullet is fired horizontally, what are the horizontal and vertical components of the initial speed?
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1 month ago
#4
(Original post by mathstutor24)
baryonplease can you delete this. We are only supposed to give hints and help, not full answers. Thanks
sorry
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1 month ago
#5
(Original post by baryon)
sorry
:-). Your vertical motion was a bit large.
Last edited by mqb2766; 1 month ago
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#6
(Original post by mathstutor24)
As mqb2766 has suggested, you need to split the motion into its horizontal and vertical components.

If the bullet is fired horizontally, what are the horizontal and vertical components of the initial speed?
(Original post by mqb2766)
What is your problem? I presume you're analysing horizontal and vertical motion seperately?
horizontally , s=100m , u=200m/s , a=0m/s/s , t=0.5s

vertically , s=1.22625m , u=0m/s , a=9.81m/s/s , t=0.5

i'm not sure how to apply the components of SUVAT to figure out the second part of the question (the angle the bullet makes with the horizontal)
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1 month ago
#7
(Original post by anon8rfy7u)
horizontally , s=100m , u=200m/s , a=0m/s/s , t=0.5s

vertically , s=1.22625m , u=0m/s , a=9.81m/s/s , t=0.5

i'm not sure how to apply the components of SUVAT to figure out the second part of the question (the angle the bullet makes with the horizontal)
Think about the distance it has travelled horizontally and vertically from its initial position to its final position (target). If you draw a horizontal line of 100m and a vertical line of 1.22625m, then a line connecting the bullet's initial position and its final position, what shape will you get?
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#8
(Original post by mathstutor24)
Think about the distance it has travelled horizontally and vertically from its initial position to its final position (target). If you draw a horizontal line of 100m and a vertical line of 1.22625m, then a line connecting the bullet's initial position and its final position, what shape will you get?
a right angled triangle?
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1 month ago
#9
(Original post by anon8rfy7u)
a right angled triangle?
Yes! So if you draw that triangle, add in the lengths of the horizontal and vertical sides, how can you calculate the angle with the horizontal?
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#10
(Original post by mathstutor24)
Yes! So if you draw that triangle, add in the lengths of the horizontal and vertical sides, how can you calculate the angle with the horizontal?
arctan ( vertical / horizontal) for angle with the horizontal ?
Last edited by anon8rfy7u; 1 month ago
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1 month ago
#11
(Original post by anon8rfy7u)
arctan (vertical / horizontal ) for angle above the horizontal ?
Yep
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#12
(Original post by mathstutor24)
Yep
that gave me an answer of 0.7025542836 degrees, i thought that would be way too small
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1 month ago
#13
(Original post by mathstutor24)
Think about the distance it has travelled horizontally and vertically from its initial position to its final position (target). If you draw a horizontal line of 100m and a vertical line of 1.22625m, then a line connecting the bullet's initial position and its final position, what shape will you get?
It's not totally clear what's meant, but I would have thought the angle the bullet makes with the horizontal should be a function of its velocity, not its position.
Last edited by DFranklin; 1 month ago
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1 month ago
#14
(Original post by DFranklin)
The angle the bullet makes with the horizontal is a function of its velocity, not its position, surely?
Sorry - yes, ofc you are correct!

I'm trying to do multiple things at once and wasn't really concentrating. No excuse for poor advice - very sorry!
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1 month ago
#15
(Original post by mathstutor24)
Sorry - yes, ofc you are correct!

I'm trying to do multiple things at once and wasn't really concentrating. No excuse for poor advice - very sorry!
If you look, you'll see I updated my response - I really don't like the way it's worded as I think the "common sense" interpretation is the one you made initially. Although I think they do want an answer using the velocity, I'd be much more certain if they'd used different wording.

Maybe I'm overthinking it.
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1 month ago
#16
(Original post by DFranklin)
If you look, you'll see I updated my response - I really don't like the way it's worded as I think the "common sense" interpretation is the one you made initially. Although I think they do want an answer using the velocity, I'd be much more certain if they'd used different wording.

Maybe I'm overthinking it.
I agree, the wording is poor. However, I do think the answer should be using velocity.

anon8rfy7u - calculate v for both components (remembering that the horizontal component of velocity remains constant throughout the motion). Then use the same method as you did with the distance to calculate the direction of motion (this will still be very small btw).
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1 month ago
#17
(Original post by anon8rfy7u)
arctan ( vertical / horizontal) for angle above the horizontal ?
Agree with the ambiguity about position or velocity angle, and presuming the "above" is a typo?

For the horizontal/vertical velocities & times in question, it would be surprising if the angle wasn't small in magnitude.
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#18
(Original post by mqb2766)
Agree with the ambiguity about position or velocity angle, and presuming the "above" is a typo?

For the horizontal/vertical velocities & times in question, it would be surprising if the angle wasn't small in magnitude.
yep, was a typo.

so would I work out the arctan of 200/4.905 ?
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1 month ago
#19
(Original post by anon8rfy7u)
yep, was a typo.

so would I work out the arctan of 200/4.905 ?

If it's direction of motion, then no. 4.95/200 (edited) or -4.95/200 if positive y is upwards.
Noting it's "negative" or pointing below the horizontal.
Last edited by mqb2766; 1 month ago
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