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

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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...

It's probably really easy for most people here! Could anyone help, please?

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...

It's probably really easy for most people here! Could anyone help, please?

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#2

(Original post by

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...

It's probably really easy for most people here! Could anyone help, please?

**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...

It's probably really easy for most people here! Could anyone help, please?

1

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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!

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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#5

(Original post by

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!

**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!

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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!)?

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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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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#11

(Original post by

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?

**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?

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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#12

(Original post by

But you can't really do that without the time

**CrisSBaader**)But you can't really do that without the time

a=10ms

^{2}

u=0

so rearrange the equationn to get t

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#13

(Original post by

Could anyone help, please?

**madele**)Could anyone help, please?

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#14

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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#15

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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#16

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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#17

(Original post by

Extra bit - For resolving vectors into components if u have the resultant force... x directions = resulantx(costheta) and direction = resultantx(sintheta)

**joshohill**)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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(Original post by

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?

**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?

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#20

use this for formulas, although if you are aqa physics AS they are in your formula sheet

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