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Constant acceleration problems
By Yiingleong Chin (P3505) on Saturday, December 30, 2000 - 02:08 am:

I have a couple of Physics questions. Hope you can help.

1. A physics student came up with a scheme to measure a buildings height. A timing mechanism is set up to measure the time needed for a lead weight dropped from the top of the building to fall the last 1.5 m before hitting the ground. Observations for a certain building show that the weight takes 0.109 s to fall this last 1.5 m. How high is the building?

2. A railcar is moving horizontally with a speed of 24m/s and decelerating at 3.65 m/s2 when a light bulb 2.55 m above the floor comes loose and drops. Where, relative to the point directly below its original position, will the bulb strike the floor?

By The Editor:

These two problems are both in situations of constant acceleration, and some combination of the following formulae is likely to be useful in these and similar problems:
s=ut+½at2
v2-u2=2as
v=u+at
s=½(u+v)t

By Carl Evans (P2080) on Sunday, December 31, 2000 - 04:23 am:

Question 1:
Acceleration is 9.8m/s, distance is 1.5m, and time taken is 0.109s. We need to find the velocity when time was started. Using s=ut+1/2at2, you can find u (the initial velocity), by substituting a, t, and s. The answer is 13.2274m/s.

Then we can find the total distance, once we have found x-1.5m. The initial velocity of the weight is 0 (because its dropped), the final velocity is 13.2274m/s, and acceleration is 9.8m/s. Using v2=u2+2as, x-1.5m is 8.9267m. Therefore total distance, and height of building, is 10.43m. All this assumes air resistance is negligible.

Carl

By Carl Evans (P2080) on Sunday, December 31, 2000 - 04:48 am:

Question 2:
First, find time taken for bulb to hit floor. Use s=ut+1/2at2, using a=9.8m/s, s=2.55m, and u=0. t=0.72139s. Then use this value of t, to find distance the train moves away from the place directly under the bulb, using s=ut+1/2at2 again, with a=-3.65, u=24, and t=0.72139. Distance is 16.364m. Therefore that is the distance directly below where the bulb was dropped to where it landed.

Carl

By Yiingleong Chin (P3505) on Tuesday, January 2, 2001 - 12:24 pm:

Thanks!