Question
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.
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.
Answers
Damon
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 :)
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 :)