# Projectile Motion Question

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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

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

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#2

(Original post by

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

**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

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#3

**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

If the bullet is fired horizontally, what are the horizontal and vertical components of the initial speed?

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#4

(Original post by

baryonplease can you delete this. We are only supposed to give hints and help, not full answers. Thanks

**mathstutor24**)baryonplease can you delete this. We are only supposed to give hints and help, not full answers. Thanks

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#5

(Original post by

sorry

**baryon**)sorry

Last edited by mqb2766; 1 month ago

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(Original post by

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?

**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

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

**mqb2766**)What is your problem? I presume you're analysing horizontal and vertical motion seperately?

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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#7

(Original post by

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)

**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)

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(Original post by

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?

**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?

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#9

(Original post by

a right angled triangle?

**anon8rfy7u**)a right angled triangle?

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(Original post by

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?

**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?

Last edited by anon8rfy7u; 1 month ago

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#11

(Original post by

arctan (vertical / horizontal ) for angle above the horizontal ?

**anon8rfy7u**)arctan (vertical / horizontal ) for angle above the horizontal ?

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(Original post by

Yep

**mathstutor24**)Yep

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#13

**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?

Last edited by DFranklin; 1 month ago

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#14

(Original post by

The angle the bullet makes with the horizontal is a function of its velocity, not its position, surely?

**DFranklin**)The angle the bullet makes with the horizontal is a function of its velocity, not its position, surely?

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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#15

(Original post by

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!

**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!

Maybe I'm overthinking it.

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#16

(Original post by

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.

**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.

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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#17

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

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(Original post by

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.

**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.

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

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#19

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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