Amusment park for physics class

[Format]

Went to an amusment park yesterday for physics class. Went on a ride called the "hellevator" where you sit down and it shoots up a certain distance then let's you free fall. My question is if the acceleration upwards occurs in the first 5 meters how do you figure out that acceleration?

 

[PRodQuanta]

Acceleration=change in velocity/time.

Format said: Went to an amusment park yesterday for physics class. Went on a ride called the "hellevator" where you sit down and it shoots up a certain distance then let's you free fall. My question is if the acceleration upwards occurs in the first 5 meters how do you figure out that acceleration?

So, All you need to know is how fast you were going in the first five meters, and how long it took you to reach your final velocity.

Punch in the numbers and whalla! !Y voy!

Paden Roder

 

[Format]

lol k i think i got it. Thx

 

[Gokul43201]

I think the real problem involves figuring out the acceleration using the maximum height reached, H. You'll find that (the acceleration),
a = g*(H-5)/5, where g = 9.8 m/s^2.

 

[minstrel]

Hi,

To find the acceleration for the first five metres, we need to know the velocity at the end of that upward acceleration. Two parts to the trip: the first five metres, and the rest of the distance up. If max height attained from take off is H, then, for the second part of the trip:

v (initial) = what we need to find out
v (final) = zero (at max height)
a = 9.8 m/s^2 [down] (if friction and air resistance can be ignored, only gravity acts)
d = (H-5) [up]

solve for v (initial) using vf^2 - vi^2 = 2ad

The v (initial) you just discovered is the v (final) for the 5 metres of acceleration at the beginning. You now use:

v (initial) = zero
v (final) = v (initial) from calculation above
d = 5.0 m [up]

and solve for acceleration using

vf^2 - vi^2 = 2ad

Hope this helps.

Minstrel