Calculating gradient with distance and speed

[Mukilab]

Homework Statement

http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_PC_2006_QP.pdf

Question 11

Homework Equations

h=v^2/g

The Attempt at a Solution

convert 36km/h into 10m/s
10 is final velocity so average should be 5m/s
h=v^2/g=5^2/10=2.5m
Using pythagoras theorem we can calculate that the other length must be roughly 50
(2.5^2+x^2=50^2)

Therefore since the gradient is the change in y over the change in x,
shouldn't the gradient just be 2.5/50=0.05? The actual answer is 0.101, they say the word 'tangent' in the markscheme but offer no explanation for it and I do not understand it.

Any help would be great, thank you very much

 

[Liquidxlax]

you have the final speed and initial speeds 10m/s and 0m/s. You have the distance travelled.

The first thing to do would to be calculate the acceleration and if you look at your kinematic equations

you'll notice the second equation fits what you have

a = v_f²/(2d)

you know that the y acceleration is gravity with a value of 9.81m/s^2

you can use trig to find the angle of the gradientsin(theta) = ?

 

[Mukilab]

you have the final speed and initial speeds 10m/s and 0m/s. You have the distance travelled.

The first thing to do would to be calculate the acceleration and if you look at your kinematic equations

you'll notice the second equation fits what you have

a = v_f²/(2d)

you know that the y acceleration is gravity with a value of 9.81m/s^2

you can use trig to find the angle of the gradient

sin(theta) = ?

I actually did use that exact formula to find acceleration but I didn't see how it could be used to find a gradient

I found a=1m/s^2

However these are acceleration, so can I still use them to find the gradient?

>Checking
sin(10/1)=0.017

using g=9.81 still gives 0.17 which is also wrong.

.__. where have I gone wrong?

Thank you very much for your help so far

The answer is 0.101 (3 sf) I realized I hadn't mentioned it so far

 

[Liquidxlax]

its been a long long time since I've done this kind of physics

the best thing to do is draw a picture it helps a lot

i drew this but i think i mucked up, that A should be in the top left of the of the large triangle because the angle i go would mean acceleration would be close to g if A was where i drew it.all i know is that i got 84.15 degrees. which i don't think is right and now this will bug me...

but yes the acceleration down the hill is 1m/s

 

[Mukilab]

its been a long long time since I've done this kind of physics

the best thing to do is draw a picture it helps a lot

i drew this but i think i mucked up, that A should be in the top left of the of the large triangle because the angle i go would mean acceleration would be close to g if A was where i drew it.


all i know is that i got 84.15 degrees. which i don't think is right and now this will bug me...

but yes the acceleration down the hill is 1m/s

The answer of the gradient is 0.101.

Can anyone remind me how gradient is related to angles?

Thank you for your time and effort, although I'm a bit confused about your diagram :S

 

[Mukilab]

Nevermind I figured it out with my friend.

If you use sin rule you can get the angle

then tan of that angle is the gradient.

I appreciate the help, thank you