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An archer fires an arrow towards a target as shown below

The diagram isn’t drawn to scale

The centre of the target is at the same height as the initial post ion of the arrow

The target is a distance of 90m from the arrow

The arrow has an initial velocity of 68ms-1 and is fired at an angle of 11 degrees to the horizontal

Air resistance has negligible effect on the motion of the arrow.

Show that the time taken for the arrow to reach its maximum height is about 1.3s.

I’m confused on how to do this at all, could someone please help

The diagram isn’t drawn to scale

The centre of the target is at the same height as the initial post ion of the arrow

The target is a distance of 90m from the arrow

The arrow has an initial velocity of 68ms-1 and is fired at an angle of 11 degrees to the horizontal

Air resistance has negligible effect on the motion of the arrow.

Show that the time taken for the arrow to reach its maximum height is about 1.3s.

I’m confused on how to do this at all, could someone please help

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#2

do you have an equation list? if so look for the equation that models h(t) (height in reference to time) and fill it with your given values, one thing you'll need to work out as well is the max height(m), which will be using sin(11)=x/90

hope this helps

hope this helps

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(Original post by

do you have an equation list? if so look for the equation that models h(t) (height in reference to time) and fill it with your given values, one thing you'll need to work out as well is the max height(m), which will be using sin(11)=x/90

hope this helps

**i.am.here**)do you have an equation list? if so look for the equation that models h(t) (height in reference to time) and fill it with your given values, one thing you'll need to work out as well is the max height(m), which will be using sin(11)=x/90

hope this helps

S =

U = 68

V = /

A = 9.81

T = ?

So you said find the max height, to do this do I do..

Sin (11) = x/90 —> x = sin(11) x 90 = 17.17

S = 1/2 x (u + v) x t —-> 17.17 = 1/2 x (68 + 0) x t —-> 17.17 = 34 x t

But then I got 17.17/34 = 0.5 which isn’t 1.3s so I’m a bit confsued

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(Original post by

Could you use suvat ?? So:

S =

U = 68

V = /

A = 9.81

T = ?

So you said find the max height, to do this do I do..

Sin (11) = x/90 —> x = sin(11) x 90 = 17.17

S = 1/2 x (u + v) x t —-> 17.17 = 1/2 x (68 + 0) x t —-> 17.17 = 34 x t

But then I got 17.17/34 = 0.5 which isn’t 1.3s so I’m a bit confsued

**Mal.singhh21**)Could you use suvat ?? So:

S =

U = 68

V = /

A = 9.81

T = ?

So you said find the max height, to do this do I do..

Sin (11) = x/90 —> x = sin(11) x 90 = 17.17

S = 1/2 x (u + v) x t —-> 17.17 = 1/2 x (68 + 0) x t —-> 17.17 = 34 x t

But then I got 17.17/34 = 0.5 which isn’t 1.3s so I’m a bit confsued

i.am.here could u also see if this is correct pls

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#5

Consider having a function of some sort. . Assume that you can find the derivative of that function. Let us call it . If you set that derivative to zero, it gives you the maximum and minimum of the function.

You already have a function which describes the height of the trajectory with time. So you can use that logic to get .

You already have a function which describes the height of the trajectory with time. So you can use that logic to get .

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ThiagoBrigido

V^2 = u^2 + 2as —-> 0^2 = 4624 + 2 x 9.81 x s

0 = 4643.62 x s

So s = 4643.62

S = u + at —> 4643.62 = 68 + (9.81 x t)

4575.62 = 9.81 x t

T = 466.42...

That’s what I got

Where have I gone wrong?

V^2 = u^2 + 2as —-> 0^2 = 4624 + 2 x 9.81 x s

0 = 4643.62 x s

So s = 4643.62

S = u + at —> 4643.62 = 68 + (9.81 x t)

4575.62 = 9.81 x t

T = 466.42...

That’s what I got

Where have I gone wrong?

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#8

(Original post by

Using two SUVAT equations to work out the vertical displacement. At the highest position, the velocity is equal to 0ms^-1. Find the maximum height using , then sub into to show time is 1.255..., therefore 1.3s.

**ThiagoBrigido**)Using two SUVAT equations to work out the vertical displacement. At the highest position, the velocity is equal to 0ms^-1. Find the maximum height using , then sub into to show time is 1.255..., therefore 1.3s.

Image from here:

https://jameskennedymonash.wordpress...-rearranged-3/

In fact, you don't even need to do calculus. You can effectively just state the vertical component of the velocity is zero and use one of the equations listed on that page website I listed. This is probably what they want you to do and both the calculus method and the substitution method give .

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#9

(Original post by

How have you arrived at your second equation? I also think you need to take more care in this problem since you have horizontal and vertical components at play.

Image from here:

https://jameskennedymonash.wordpress...-rearranged-3/

https://www.physicsclassroom.com/cla...s-Problem-Solv

In fact, you don't even need to do calculus. You can effectively just state the vertical component of the velocity is zero and use one of the equations listed on that page website I listed. This is probably what they want you to do and both the calculus method and the substitution method give .

**0le**)How have you arrived at your second equation? I also think you need to take more care in this problem since you have horizontal and vertical components at play.

Image from here:

https://jameskennedymonash.wordpress...-rearranged-3/

https://www.physicsclassroom.com/cla...s-Problem-Solv

In fact, you don't even need to do calculus. You can effectively just state the vertical component of the velocity is zero and use one of the equations listed on that page website I listed. This is probably what they want you to do and both the calculus method and the substitution method give .

I should have double checked it. , sure is what he needs.

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0le ThiagoBrigido

So are you both saying to do this:

Horizontal component= 68 x cos(11) = 66.75064847

Vertical component = 68 x sin(11) = 12.97501169

Then 0le the website you have said use (V_fy = V_iy + a_y x t)

So (12.975...) = (66.75...) + (-9.81) x t

So -53.775... = -9.81 x t

So t = -5.48? But that’s wrong so idk what I’ve done wrong here

Or do u mean:

Final velocity = 0

Initial velocity = 68

Distance = 90m

Angle = 11

So do v = u + at

0 = 68 + 9.81 x t

-68 = 9.81 x t

-68 / 9.81 = -6.9 ???

Idk what I’ve done wrong

So are you both saying to do this:

Horizontal component= 68 x cos(11) = 66.75064847

Vertical component = 68 x sin(11) = 12.97501169

Then 0le the website you have said use (V_fy = V_iy + a_y x t)

So (12.975...) = (66.75...) + (-9.81) x t

So -53.775... = -9.81 x t

So t = -5.48? But that’s wrong so idk what I’ve done wrong here

Or do u mean:

Final velocity = 0

Initial velocity = 68

Distance = 90m

Angle = 11

So do v = u + at

0 = 68 + 9.81 x t

-68 = 9.81 x t

-68 / 9.81 = -6.9 ???

Idk what I’ve done wrong

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#11

(Original post by

0le ThiagoBrigido

So are you both saying to do this:

Horizontal component= 68 x cos(11) = 66.75064847

Vertical component = 68 x sin(11) = 12.97501169

Then 0le the website you have said use (V_fy = V_iy + a_y x t)

So (12.975...) = (66.75...) + (-9.81) x t

So -53.775... = -9.81 x t

So t = -5.48? But that’s wrong so idk what I’ve done wrong here

**Mal.singhh21**)0le ThiagoBrigido

So are you both saying to do this:

Horizontal component= 68 x cos(11) = 66.75064847

Vertical component = 68 x sin(11) = 12.97501169

Then 0le the website you have said use (V_fy = V_iy + a_y x t)

So (12.975...) = (66.75...) + (-9.81) x t

So -53.775... = -9.81 x t

So t = -5.48? But that’s wrong so idk what I’ve done wrong here

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#12

(Original post by

An archer fires an arrow towards a target as shown below

The diagram isn’t drawn to scale

The centre of the target is at the same height as the initial post ion of the arrow

The target is a distance of 90m from the arrow

The arrow has an initial velocity of 68ms-1 and is fired at an angle of 11 degrees to the horizontal

Air resistance has negligible effect on the motion of the arrow.

Show that the time taken for the arrow to reach its maximum height is about 1.3s.

I’m confused on how to do this at all, could someone please help

**Mal.singhh21**)An archer fires an arrow towards a target as shown below

The diagram isn’t drawn to scale

The centre of the target is at the same height as the initial post ion of the arrow

The target is a distance of 90m from the arrow

The arrow has an initial velocity of 68ms-1 and is fired at an angle of 11 degrees to the horizontal

Air resistance has negligible effect on the motion of the arrow.

Show that the time taken for the arrow to reach its maximum height is about 1.3s.

I’m confused on how to do this at all, could someone please help

Last edited by S9123; 3 months ago

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#13

(Original post by

You don't need any suvats - just the s=d/t equation. Find the horizontal component of velocity by 68cos(11) and then do the following calculcation -> t=d/s = 90/68cos(11) = 1.348....s.

I hope that helps

**S9123**)You don't need any suvats - just the s=d/t equation. Find the horizontal component of velocity by 68cos(11) and then do the following calculcation -> t=d/s = 90/68cos(11) = 1.348....s.

I hope that helps

EDIT: Another problem with that method is that the arrow travels greater than because it travels along a curve (i.e. imagine that it doesn't travel along the fastest route but takes a detour, therefore having to cover more ground).

Last edited by 0le; 3 months ago

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0le

So I know that Vfy = final velocity and Vfi = initial velocity

So is Vfy = 0 and Vfi = 68

So is it

68 = 0 x (-9.81) x t

So t = 6.93?

I don’t know what I’ve done

In the website you gave thsy used the horizontal and vertical components so I’m not sure what I’ve done

So I know that Vfy = final velocity and Vfi = initial velocity

So is Vfy = 0 and Vfi = 68

So is it

68 = 0 x (-9.81) x t

So t = 6.93?

I don’t know what I’ve done

In the website you gave thsy used the horizontal and vertical components so I’m not sure what I’ve done

Last edited by Mal.singhh21; 3 months ago

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Omds I think I’ve got it!!! 0le sorry ignore the other post I thought about it again after and got this

So is it this:

Horizontal component = 0

Vertical = 12.97...

12.97... = (-9.81) x t

So t = 12.97... / (-9.81) = -1.322629969

???

So is it the horizontal component that I state is zero

So is it this:

Horizontal component = 0

Vertical = 12.97...

12.97... = (-9.81) x t

So t = 12.97... / (-9.81) = -1.322629969

???

So is it the horizontal component that I state is zero

Last edited by Mal.singhh21; 3 months ago

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#16

(Original post by

0le

So I know that Vfy = final velocity and Vfi = initial velocity

So is Vfy = 0 and Vfi = 68

**Mal.singhh21**)0le

So I know that Vfy = final velocity and Vfi = initial velocity

So is Vfy = 0 and Vfi = 68

It clearly says "vert.". What do you think that means? So what do you think should be? Do you also understand why ?

So is it

**68 = 0 x (-9.81) x t**

So t = 6.93?

I don’t know what I’ve done

Last edited by 0le; 3 months ago

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