Passengers on the Giant Drop, a free-fall ride at Six Flags Great America, sit i
ID: 1657858 • Letter: P
Question
Passengers on the Giant Drop, a free-fall ride at Six Flags Great America, sit in cars that are raised to the top of a tower. The cars are then released for 2.6 s of free fall, after which the brakes kick in. The brakes take 1.0 s to bring the passengers to rest.
a) How fast are the passengers moving at the end of the speeding up phase of the ride? = 25.48 m/s
b) What is the average acceleration (magnitude and direction) over the breaking phase of the ride? = 25.48 m/s^2
* c) Given these numbers, what is the minimum possible height of the tower? = ?
Explanation / Answer
A) During the free falling portion of the motion, the acceler-action is -g=-9.8 m/s2
v = u + a1 ( tf - ti)
v = 0 - g ( 2.6)
v = 25.48 m/s
b) During the stopping portion of the motion, we can use the same equation to find the unknown acceleration
now
v = u + a2 ( tf-ti)
0 = (25.48) - a2( 1)
a2 = 25.48 m/s2
c) To find the net displacement during the two motions, we need to separately find the displacements during each part
s = ut + 1/2 a1 t2
s1 = 0 + 1/2 (-9.81) (2.6)2
s1 = -33.15m
now s2
s2 = ut + 1/2 a2t2
s2 = (-26) x 1 + 1/2 g (1)2
s2 = - 21.09 m
so total Snet = s1+ s2 = -33.15 - 21.09
= -54.25 m
Since the ride falls 54.25 meter, the tower had better be at least that hight
answer