The ride drops riders from the top of a tall tower and let's them free fall for a short time before bringing them to a stop. For the riders' safety, you must make sure the riders do not experience an acceleration greate than 2g. Assume the riders have a constant acceleration while they slow down and come to a stop. If the riders are to free fall for 3.5 seconds, what is the minimum height of the tower?
g=9.8 and 2(9.8) = 19.6m/s^2, so the acceleration has to be less than 19.6m/s^2. Velocity = 19.6t, and speed(position) = (19.6/2)t^2, with t = 3.5s. I was expecting to just substitute t = 3.5 but it's not giving me the correct answer.
The answer given was 90.0m.
during free fall for 3.5 s:
v = g t = 9.8*3.5 = 34.3 m/s when we pt on the brakes
h = (1/2) g t^2 = 4.9 (3.5)^2 = 60 meters so far, now we stop
we are going at 34.3 m/s and slow down at 2 g
0 = Vi - 2 g t
so
t = 34.3 /19.6 = 1.75 seconds braking
h = Vi t - (1/2)(2*9.8) t^2
= 34.3 (1.75) - 9.8 (1.75)^2
= 60 - 30 = 30
(of course it will take half the distance we did in fee fall since we are braking at 2 g)
60 + 30 = 90 meters high
( I would make it a little higher if I were you :)
To determine the minimum height of the tower, we need to find the distance traveled by the riders during the free fall and equate it to the height of the tower.
Using the equation for displacement under constant acceleration:
s = ut + (1/2)at^2
Since the riders start from rest (u = 0), the equation simplifies to:
s = (1/2)at^2
We are given that the riders free fall for 3.5 seconds (t = 3.5s), and the maximum acceptable acceleration is 19.6m/s^2 (2g). Substituting these values into the equation, we have:
s = (1/2)(19.6)(3.5)^2
s = 0.5 * 19.6 * 12.25
s = 115.42m
Therefore, the minimum height of the tower should be at least 115.42 meters to ensure that the riders do not exceed an acceleration greater than 2g.
It appears that the answer provided (90.0m) may not be correct. The correct answer should be 115.42m.