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

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#1
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
4 years ago
#2
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?
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
1
#3
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!
0
4 years ago
#4
s=ut + at2/2

I get 31.95m/s assuming g=10m/s
0
4 years ago
#5
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.
0
4 years ago
#6
S isn't 105 with g.

It's 54
0
4 years ago
#7
(Original post by Maker)
s=ut + at2/2
But you can't really do that without the time
0
4 years ago
#8
(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.
0
#9
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!)?
0
#10
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?
0
4 years ago
#11
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
0
4 years ago
#12
But you can't really do that without the time
S=54m
a=10ms2
u=0
so rearrange the equationn to get t
0
4 years ago
#13
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.
1
4 years ago
#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?
0
4 years ago
#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.
0
4 years ago
#16
Extra bit - For resolving vectors into components if u have the resultant force... x directions = resulantx(costheta) and direction = resultantx(sintheta)
0
4 years ago
#17
(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.
0
#18
Oh no, the answer I have worked out is 110 m/s
0
#19
(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.
0
4 years ago
#20
use this for formulas, although if you are aqa physics AS they are in your formula sheet
0
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