Question

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?

Answer 1

The velocity is
`v_h = 19.2 \ m/s`

Step-by-step explanation:

From the question we are told that

The speed of the roller coaster at ground level is
`v = 26 \ m/s`

Generally we can define the roller coaster speed at ground level using the an equation of motion as

`v^2 = u^2 + 2 g s`

u is zero given that the roller coaster started from rest

So

`26^2 = 0 + 2 * g * s`

So

`s = (26^2)/( 2 * g )`

=>
`s = 37.6 \ m`

Now the displacement half way is mathematically represented as

`s_(h) = (37.6)/(2)`

`s_(h) = 18.8 \ m`

So

`v_h ^2 = u^2 + 2 * g * s_h`

Where
`v_h` is the velocity at the half way point

=>
`v_h = √( 0 + 2 * 9.8 * 18.8 )`

=>
`v_h = 19.2 \ m/s`