Question

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 28 m/s when it reaches ground level. part a what was its speed when its height was half that of its starting point?

Answer 1

Let the distance be s
initial velocity was u = 0m/s
final velocity was v = 28m/s
Since no acceleration is provided, only acceleration here is due to gravity. So here, a = 9.81m/s2

Using Newtons motion equation,

v^2 = u^2 + 2as
28^2 = 0 + 2 x 9.81 x s
s = 784/19.62
s = 81.55m
s/2 = 39.96m

Now,
same equation, to find v at s/2
v^2 = 0 + 2 x 9.81 x 39.96
v^2 = 784.0152
v = 28m/s