1. A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 200m. Disregarding friction, what is the velocity of the bobsled at the bottom of the hill? (g=9.81 m/s^2)

Question

jziggy · Answer

This can be solved easily with conservation of energy, at the top of the hill and at the bottom of the hill it will have the same energy.

jziggy · Answer

So, m*h*g + 0 = 0 + mv^2
h*g = v^2

jziggy · Answer

v = sqrt(h*g) and then you just need to input the numbers

anonymous · Answer

ok one sec let me fill them in

anonymous · Answer

(1/2) m v^2 = m h g  note the 2

v = sqrt (2 h g)

jziggy · Answer

oooh derp

jziggy · Answer

It's been a bit too long since i've done this >.<

jziggy · Answer

but yes it's v = sqrt(2*h*g)

anonymous · Answer

so sqrt(2*200*9.81) then multiply right?

jziggy · Answer

yes

anonymous · Answer

the answer would be 62.64?

jziggy · Answer

Depends on how many significant figures you need, since they gave gravity with 3 I'd usually use 3 sig figs in my answer as well.

anonymous · Answer

oh okay can you help me with more?

jziggy · Answer

I can in about 30 minutes if you still need it

anonymous · Answer

okay ill have a question posted on here so you can help me when you can thank you