Anew ride being built at an amusement park includes a vertical drop of 126.5 meters. starting from rest, the ride vertically drops that distance before the track curves forward. if friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?

17.60 m/s
24.90 m/s
49.79 m/s
70.42 m/s

Speed at the bottom of the roller coaster = 49.79 m/s

Explanation:

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically drops that distance before the track curves forward.

We have to find the speed at the bottom.

Here the gravitational energy fully converts to kinetic energy, so we equate it.

Gravitational energy =

Kinetic energy =

=

= 2479.4

Velocity, v = 49.79 m/s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

what is the actual question though?