Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?

Question

Acceleration due to gravity on Earth is 9.81 m/s2. This means that any object on Earth in "free fall" will have an acceleration of 9.81 m/s2. Also, objects in free fall have an initial velocity of zero. Conversely, any object that is thrown or shot straight up will have a "final" velocity of zero at the top of its motion after experiencing a negative acceleration due to gravity of, you guessed it, 9.81 m/s2. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?

a) Final Velocity:

b) Distance:

Answer 1

The most important step is first to define our initial and final positions, then list the known information at those points. We also have to define what direction is positive/negative, and make sure any vectors that point in those directions have the correct sign. 3 is the magic number; if we can find 3 pieces of info, we can solve. This method will solve you any kinematics problem.

a) initial position: at the top

final position: after falling 2.60 seconds

Define up as positive, down as negative

knowns (given)

change in t = 2.60 seconds

initial velocity at top vi = 0 m/s

acceleration a = -9.81m/s^2 (negative since it’s pulling down)

unknown that we want to find: final velocity vf

Now that we have 3 pieces of information, we are going to look at our kinematics equations. We want to find the equation that only has the 3 knowns and 1 unknown in it, and no other variables - in this case, it can only contain vi, t, a, and vf

kinematics equations

vf = vi + at

vf^2 = vi^2 + 2*a* change in position

change in position = Vi * t + 1/2a * t^2

The equation that has everything we need is the first one. Plug in the knowns to find vf:

vf = 0 + (-9.81)(2.60)

vf = -25.506 m/s This is the velocity. The - sign shows that we are moving downward

(Most teachers would accept if you gave this number as positive for a final answer, but if you use it as positive in the equations you will get the wrong answer so be careful!)

b) Next we are finding the distance fallen. Let’s start with the same knowns from part a and this time add on our new known, vf.

knowns (given)

change in t = 2.60 seconds

initial velocity at top vi = 0 m/s

acceleration a = -9.81m/s^2 (negative since it’s pointing down)

vf after 2.60 seconds = -25.506 m/s negative since it’s falling down)

unknown that we want to find: distance

Notice that distance is not an option in the kinematics equations, but we can calculate it through change in position. Distance is the scalar form of change in position (the same way that speed is the scalar form of velocity). Distance is change in position without caring about the direction it’s going - so we will find the change in position and give the answer positive, regardless of the sign we get. Let’s use the 3rd equation:

change in position = Vi *t + 1/2a*t^2

change in position = 0 * 2.60 + 1/2(-9.81)(2.60)^2

change in position = 0 + 1/2(-9.81)(6.76)

change in position = 0 + -33.16

change in position = -33.16 m (this is negative because we are falling down)

distance = +33.16 m