Physics (energy)

Wagon in the figure can roll down the hill without friction. When the outpot speed (Vo) is 0 m/s and the end speed (v) is 4m/s

How high is the hill?

$E=\frac{m}{2}(v^2-v_o^2)=mgh$

$v^2-v_0^2=2gh$

$(4\frac{m}{s})^2=2\cdot 9.81\frac{m}{s^2}\cdot h$

$h=16\cdot \frac{m^2}{s^2}/2\cdot9.81\cdot\frac{m}{s^2}$ 

$h=\frac{16\frac{m^2}{s^2}}{2\cdot 9.81\frac{m}{s^2}}$

$h=0.815\ m$

$The \ hill \ is \ 0.815 \ m \ high.$

  !

Use conservation of energy:

potential energy at the top = m*g*h   where m is mass, g is acceleration of gravity and h is height.

kinetic energy at the bottom = (1/2)mv²    where v is velocity

Equating the two we get

gh = (1/2)v²    or   h = (1/2)v²/g

You can crunch the numbers!

Looks like asinus has already crunched the numbers for you!