How do I find speed with only distance and not time? Watch

madele
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I didn't choose to do Physics - it's just part of my course and I am really poor at it.

I have this question that I'm really confused about. Here it is:

A movie stunt driver on a motorcycle speeds horizontally off a 54m high cliff. How fast must the motorcycle leave the cliff‐top if it is to land on the level ground below at a distance of 105m from the base of the cliff?

I thought that to work out a speed you need another value... I thought maybe I need to use Pythagoras but I don't actually know how that would help... :confused:

It's probably really easy for most people here! Could anyone help, please?
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CrisSBaader
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(Original post by madele)
I didn't choose to do Physics - it's just part of my course and I am really poor at it.

I have this question that I'm really confused about. Here it is:

A movie stunt driver on a motorcycle speeds horizontally off a 54m high cliff. How fast must the motorcycle leave the cliff‐top if it is to land on the level ground below at a distance of 105m from the base of the cliff?

I thought that to work out a speed you need another value... I thought maybe I need to use Pythagoras but I don't actually know how that would help... :confused:

It's probably really easy for most people here! Could anyone help, please?
Ok, so you need to use the equations of motion, if you have s(distance)=105, a(acceleration due to gravity)=9.81, V(final velocity)=0, you could use V^2=U^2+2as
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madele
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I didn't think we knew the final velocity to be zero hence not using this formula I have v^2 = u^2 + 2gs.

So are you saying then I am supposed to presume the motorcycle comes to a stop? I never know in these questions!
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Maker
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s=ut + at2/2

I get 31.95m/s assuming g=10m/s
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CrisSBaader
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(Original post by madele)
I didn't think we knew the final velocity to be zero hence not using this formula I have v^2 = u^2 + 2gs.

So are you saying then I am supposed to presume the motorcycle comes to a stop? I never know in these questions!
I would say so, I don't really think that there is another way of doing it without overcomplicating it.
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L'Evil Fish
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S isn't 105 with g.

It's 54
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CrisSBaader
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(Original post by Maker)
s=ut + at2/2
But you can't really do that without the time
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CrisSBaader
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(Original post by L'Evil Fish)
S isn't 105 with g.

It's 54
Upss...
This is why you should always read the questions properly.
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madele
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OK thank you

However I'm still stuck

I don't know what s should be. Should it be 54 (as the formula is for vertical motion only) or was I right to think of using Pythagoras's Theorem? Or something else (probably!)?
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madele
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Or rather do I work the vertical motion out (I've come up with -32.5 m/s) and then have to do something else to work it out completely?
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L'Evil Fish
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(Original post by madele)
Or rather do I work the vertical motion out (I've come up with -32.5 m/s) and then have to do something else to work it out completely?
Find t from vertical motion

Then s = ut

You will have s and t, so can find u

The s = ut is from s = ut + 1/2 at^2 but a is zero horizontally
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Maker
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(Original post by CrisSBaader)
But you can't really do that without the time
S=54m
a=10ms2
u=0
so rearrange the equationn to get t
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Good bloke
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(Original post by madele)
Could anyone help, please?
The answer comes in two stages. First work out the time the bike is in the air (it falls from a height of 54m under gravity). Then use that time to work out the rest of the problem.
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joshohill
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Have you covered vectors in your physics course so far, if not that is probably why you are struggling. Remember that speed is conserved in x direction so that doesnt change and in the y direction it does change. Find the two components (x&y) to help?
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Maker
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You need to find the time it takes for the bike to fall 54m then use the time to work out the horizontal speed of the bike.
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joshohill
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Extra bit - For resolving vectors into components if u have the resultant force... x directions = resulantx(costheta) and direction = resultantx(sintheta)
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Good bloke
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(Original post by joshohill)
Extra bit - For resolving vectors into components if u have the resultant force... x directions = resulantx(costheta) and direction = resultantx(sintheta)
There are no forces or vectors in the calculations for this problem. It is simply a Newtonian equation or two.
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madele
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Oh no, the answer I have worked out is 110 m/s
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madele
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(Original post by joshohill)
Have you covered vectors in your physics course so far, if not that is probably why you are struggling. Remember that speed is conserved in x direction so that doesnt change and in the y direction it does change. Find the two components (x&y) to help?
Yes but unfortunately I really struggle to get my head round it.
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joshohill
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use this for formulas, although if you are aqa physics AS they are in your formula sheet Name:  902267042_orig.png
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