## Homework Statement

You ride your bicycle down a hill. Your total mass is 54 kg. At the top of the hill, your
speed is 5 m/s. The hill is 1 m high and 101 m long. what is
your speed at the bottom of the hill?

## The Attempt at a Solution

PE= mgh
(54)(9.8)(10)
= 5292

find the degree of the hill i think using tangent?
tan (10/100) ??

then solves for v

KE = 1/2 mv^2

Ik its something along those lines but im a bit confused with the hill (do i use tangent, etc.) thank you very much, any help as soon as they possibly can is very much appreciated. thanks :-)

Kurdt
Staff Emeritus
Gold Member
Well what happens is that at the bottom of the hill all the potential energy will be converted to kinetic energy (disregarding friction and air resistance). What you need to do is set the potential and kinetic energy equal to each other and solve for v. Remember to add on the initial 5m/s.

To find the angle, it depends on if the 100m relates to the hypotenuse or the distance from the start of the hill to the bottom. model the hill as a right angled triangle and remember your trigonometric ratios.

KE = 1/2 mv^2
5292 = 1/2 (54)v^2
and solve for V?
and then just add on the 5m/s to that?

For the length of the hill, i am guessing that it would be referring to the hypotenuse (because that would be the length that the bike travels).

Kurdt
Staff Emeritus
Gold Member
Ok so what trigonometric function do you know that involves the opposite side (10m) and the hypotenuse (100m).

And yes just solve for v and add the 5m/s on.

KE = 1/2 mv^2
5292 = 1/2 (54)v^2
and solve for V?
V=196 + 5 = 201
but what do i do with the hill? it would be Tan-1 (10/100)?

Kurdt
Staff Emeritus
Gold Member
Remember:

$$\sin(\theta) = O/H \cos(\theta) = A/H \tan(\theta) = O/A$$

where O = opposite, A = Adjacent and H = hypotenuse.

Sin = o/h
10/100
= .0017
would that b right? but how do i figure it out now, the speed at the bottom of the hill, once ive got those numbers? thank you i really appreciate all of the help!

Last edited:
Kurdt
Staff Emeritus
Gold Member
KE = 1/2 mv^2
5292 = 1/2 (54)v^2
and solve for V?
V=196 + 5 = 201

You forgot to take the square root.

You want to take sin-1 (10/100)

okay, now when i get both of those numbers what do i do with the angle?

thanks again
for taking the time to help me

Kurdt
Staff Emeritus
Gold Member
Oh right. Nothing because you have used conservation of energy instead of kinematic equations so you don't really need the angle. I thought it was part of the question you'd neglected to post.

so the answer would just be the square root of 201 which would be 14.177

Kurdt
Staff Emeritus
Gold Member
No the square root of 196 and then add the five.

Okay, thanks very much. just one more quick question...Why don't I do anything with the hill (as in tangent) because it is going down an incline, wouldn't that have an effect on the speed (speed it up)? sorry for all of the questions

Last edited:
Kurdt
Staff Emeritus