Question

A popular ride at an amusment park lifts
customers up to a height of 50 m and then
drops them threw a displacement of 50 m
before slowing them to a stop. How fast
are the customers going at the 50 m
mark?

Answer 1

Answer:

$31.32\ m/s$

Explanation:

$We\ are\ given\ that:\\Height\ to\ which\ there're\ lifted=50m\\Displacement\ during\ the\ descent=50m\\Now,\\In\ order\ to\ find\ the\ velocity\ of\ the\ customers\ at\ 50\ m,\\We\ can\ use\ the\ Third\ Equation\ Of\ Motion,\ which is:\\2as=v^2-u^2\\As\ we\ know\ that,\\Acceleration\ due\ to\ gravity=9.81\ m/s^2\ or\ roughly\ 10\ m/s^2\\Displacement=50\ m\\Initial\ velocity=0\ m/s^2\\ [As\ they\ stop\ when\ they\ reach\ the\ maximum\ height\ of\ 50\ m\\ and\ begin\ their\ descent]$

$By\ reconstructing\ the\ Third\ Equation\ Of\ Motion,\ we\ have:\\2gs=v^2\\Hence,\\v^2=2*9.81*50 \\v^2=981\ m^2/s^2 \\v=\sqrt{981\ m^2/s^2} \\v=31.32\ m/s$