Mal.singhh21
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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
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i.am.here
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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
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Mal.singhh21
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(Original post by 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
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
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Mal.singhh21
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(Original post by 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
0le do you know how to do this
i.am.here could u also see if this is correct pls
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0le
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Consider having a function of some sort. y = f(t). Assume that you can find the derivative of that function. Let us call it \frac{dy}{dt} = f'(t). 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 \mathrm{t} \sim 1.323\mathrm{s}.
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Mal.singhh21
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Sorry I still understand? 0le

Can you explain it a bit more please
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Mal.singhh21
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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?
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(Original post by 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  V^2=u^2+2aS, then sub into  S = u + at to show time is 1.255..., therefore 1.3s.
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
Image from here:
https://jameskennedymonash.wordpress...-rearranged-3/


(Original post by Mal.singhh21)
Sorry I still understand? 0le

Can you explain it a bit more please
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 \mathrm{t} = 1.322\mathrm{s}.
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ThiagoBrigido
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(Original post by 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
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 \mathrm{t} = 1.322\mathrm{s}.
You're right,
I should have double checked it. :banghead:, sure  V= u + at is what he needs.
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Mal.singhh21
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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
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(Original post by 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
The formula you selected is correct. But you did not pay enough attention to what each of the variables refer to. Have another think about what are the variables \mathrm{V}_{\mathrm{f,y}} and \mathrm{V}_{\mathrm{i,y}}.
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S9123
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(Original post by 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
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Last edited by S9123; 3 months ago
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0le
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(Original post by 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
That does not seem correct. You need to calculate the time at the highest point in the trajectory, which is probably not 90 \mathrm{m}.

EDIT: Another problem with that method is that the arrow travels greater than 90 \mathrm{m} 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).
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Mal.singhh21
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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
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Mal.singhh21
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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
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0le
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(Original post by Mal.singhh21)
0le

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

So is Vfy = 0 and Vfi = 68
Well you've now written that the final component of the velocity in the vertical direction is zero. Whether you've understood why I am not sure. Take a look at the screenshot of the website:

Image

It clearly says "vert.". What do you think that means? So what do you think \mathrm{V}_{\mathrm{i,y}} should be? Do you also understand why \mathrm{V}_{\mathrm{f,y}} = 0?


So is it

68 = 0 x (-9.81) x t

So t = 6.93?

I don’t know what I’ve done
I don't know where you got this formula that I've indicated in bold. You previously wrote down the correct formula but this one, I have no idea where it has come from. I guess you've made a typo. However just in case you have not, if you multiply anything with zero the result is always zero. Even if that was a typo, you substituted your values in the equation incorrectly anyway.
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