Question

a 51 kg student is in a 8 kg sled at rest at the top of a hill. the student slides down the hill and reaches a speed of 116 km/h at the bottom. How high is the hill? express as m and round off to one decimal place

Answer 1

Solution.

$m_1=51kg;$

$m_2=8kg;$

$\upsilon=116km/h=32.2m/s;$

$\dfrac{(m_1+m_2)\upsilon^2}{2}=(m_1+m_2)gh;$

$\dfrac{\upsilon^2}{2}=gh;$

$h=\dfrac{\upsilon^2}{2g};$

$h=\dfrac{(32.2m/s)^2}{2\sdot9.81m/s^2}=52.8m;$

Answer: $h=52.8m.$