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Projectile Motion Question

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(edited 2 years ago)
Reply 1
Avatar for mqb2766
mqb2766
21
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

thanks in advance


What is your problem? I presume you're analysing horizontal and vertical motion seperately?
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

thanks in advance

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?
Reply 3
Avatar for baryon
baryon
6
Original post by mathstutor24
@baryonplease can you delete this. We are only supposed to give hints and help, not full answers. Thanks :smile:

sorry
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by baryon
sorry

:-). Your vertical motion was a bit large.
(edited 3 years ago)
Reply 5
Avatar for anon8rfy7u
anon8rfy7u
OP
8
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)
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?
Reply 7
Avatar for anon8rfy7u
anon8rfy7u
OP
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?
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?
Reply 9
Avatar for anon8rfy7u
anon8rfy7u
OP
8
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 ?
(edited 3 years ago)
Original post by anon8rfy7u
arctan (vertical / horizontal ) for angle above the horizontal ?

Yep :smile:
Reply 11
Avatar for anon8rfy7u
anon8rfy7u
OP
8
Original post by mathstutor24
Yep :smile:

that gave me an answer of 0.7025542836 degrees, i thought that would be way too small
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.
(edited 3 years ago)
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!
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.
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).
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.
Reply 17
Avatar for anon8rfy7u
anon8rfy7u
OP
8
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 ?
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.
(edited 3 years ago)

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